JEE Matrices Questions

Given a 3×3 matrix A with determinant |A| = 1/2 and trace(A) = 3. We need to find B = adj(adj(2A)) and then compute |B| + trace(B).

Recall the properties for n×n matrices (here n=3):

• For scalar k and matrix M, |kM| = kn |M|.
• adj(kM) = kn-1 adj(M).
• For invertible M, adj(adj M) = |M|n-2 M.
• trace(kM) = k · trace(M).

Start by computing |2A|:

$$|2A| = 2^3 |A| = 8 \times \frac{1}{2} = 4$$

Now compute adj(2A):

$$\text{adj}(2A) = 2^{3-1} \text{adj}(A) = 2^2 \text{adj}(A) = 4 \text{adj}(A)$$

Next, compute adj(adj(2A)) = adj(4 adj(A)):

$$\text{adj}(4 \text{adj}(A)) = 4^{3-1} \text{adj}(\text{adj}(A)) = 4^2 \text{adj}(\text{adj}(A)) = 16 \text{adj}(\text{adj}(A))$$

Thus, B = 16 adj(adj(A)).

Since A is invertible (|A| ≠ 0), apply the identity adj(adj A) = |A|n-2 A:

$$\text{adj}(\text{adj}(A)) = |A|^{3-2} A = \left(\frac{1}{2}\right)^1 A = \frac{1}{2} A$$

Substitute back:

$$B = 16 \times \frac{1}{2} A = 8A$$

Now compute |B|:

$$|B| = |8A| = 8^3 |A| = 512 \times \frac{1}{2} = 256$$

Compute trace(B):

$$\text{trace}(B) = \text{trace}(8A) = 8 \times \text{trace}(A) = 8 \times 3 = 24$$

Finally, add |B| and trace(B):

$$|B| + \text{trace}(B) = 256 + 24 = 280$$

The value is 280, which corresponds to option D.

We are given a $$3 \times 3$$ matrix $$A$$ with $$|A| = 5$$ and we must evaluate
$$|\,2\,\operatorname{adj}\!\bigl(3A \,\operatorname{adj}(2A)\bigr)|$$.

Step 1: Pull out the scalar “2”.
For a $$3 \times 3$$ matrix $$M$$, $$|kM| = k^{3}\,|M|$$. Hence
$$|\,2\,\operatorname{adj}(3A \operatorname{adj}(2A))| = 2^{3}\,|\operatorname{adj}(3A \operatorname{adj}(2A))|.$$

Step 2: Determinant of an adjugate.
For any $$3 \times 3$$ matrix $$C$$, $$|\operatorname{adj}C| = |C|^{3-1} = |C|^{2}$$.
Thus
$$|\operatorname{adj}(3A \operatorname{adj}(2A))| = \bigl|\,3A \operatorname{adj}(2A)\bigr|^{2}.$$

Step 3: Expand the determinant inside.
Using $$|PQ| = |P|\,|Q|$$,
$$\bigl|\,3A \operatorname{adj}(2A)\bigr| = |\,3A| \; \cdot \; |\operatorname{adj}(2A)|.$$

Step 4: Evaluate $$|\,3A|$$.
Again, $$|kA| = k^{3}|A| \implies |\,3A| = 3^{3}\,|A| = 27 \times 5 = 3^{3}\cdot 5.$$

Step 5: Evaluate $$|\operatorname{adj}(2A)|$$.
First find $$|\,2A| = 2^{3}|A| = 8 \times 5 = 2^{3}\cdot 5.$}
Then $$|\operatorname{adj}(2A)| = |\,2A|^{2} = (2^{3}$$\cdot$$ 5)^{2} = 2^{6}$$\cdot$$ 5^{2}.$$

Step 6: Combine results of Steps 4 and 5.
$$\bigl|\,3A \operatorname{adj}(2A)\bigr| = (3^{3}$$\cdot$$ 5)\,(2^{6}$$\cdot$$ 5^{2}) = 2^{6}\,3^{3}\,5^{3}.$$

Step 7: Square the determinant (per Step 2).
$$\bigl|\,3A \operatorname{adj}(2A)\bigr|^{2} = \bigl(2^{6}\,3^{3}\,5^{3}\bigr)^{2} = 2^{12}\,3^{6}\,5^{6}.$$

Step 8: Multiply by the outer factor $$2^{3}$$ (from Step 1).
Final value:
$$|\,2\,\operatorname{adj}(3A \operatorname{adj}(2A))| = 2^{3}\,(2^{12}\,3^{6}\,5^{6}) = 2^{15}\,3^{6}\,5^{6}.$$

Step 9: Identify the exponents.
$$$$\alpha$$ = 15,\; $$\beta$$ = 6,\; $$\gamma$$ = 6 \;\;\Longrightarrow\;\; $$\alpha + \beta + \gamma$$ = 15 + 6 + 6 = 27.$$

Hence the required sum is $$27$$, which corresponds to Option C.

We need to find the sum of diagonal elements of $$C = P^T B^{10} P$$, where $$B = PAP^T$$.

Since $$P$$ is a rotation matrix and thus $$P^T = P^{-1}$$, we have $$B = PAP^T \implies B^{10} = PA^{10}P^T$$. This follows because $$B^2 = (PAP^T)(PAP^T) = PA(P^TP)AP^T = PA^2P^T$$ (since $$P^TP = I$$), and by induction $$B^n = PA^nP^T$$.

Substituting into the expression for $$C$$ gives $$C = P^T B^{10} P = P^T(PA^{10}P^T)P = (P^TP)A^{10}(P^TP) = A^{10}$$, so that $$C = A^{10}$$.

Matrix $$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$$ is upper triangular, and for such matrices $$A^n$$ remains upper triangular with diagonal elements equal to the $$n$$th powers of the original diagonal elements. Hence the diagonal entries of $$A^{10}$$ are $$\left(\frac{1}{\sqrt{2}}\right)^{10}$$ and $$1^{10}$$.

Therefore, the trace of $$C$$ is $$\text{tr}(C) = \left(\frac{1}{\sqrt{2}}\right)^{10} + 1 = \frac{1}{2^5} + 1 = \frac{1}{32} + 1 = \frac{33}{32}$$, and writing this as $$\frac{m}{n}$$ gives $$\frac{m}{n} = \frac{33}{32}$$ with $$\gcd(33,32)=1$$. It follows that $$m+n = 33+32 = 65$$.

Therefore, the correct answer is Option (3): 65.

Let $$A = [a_{ij}]$$ be a $$2 \times 2$$ matrix with entries either 0 or 1, and let us find the probability that $$A$$ is invertible.

A $$2 \times 2$$ matrix has 4 entries, each of which can be 0 or 1, so the total number of such matrices is $$2^4 = 16$$.

Since $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ is invertible exactly when $$\det(A) = a_{11}a_{22} - a_{12}a_{21} \neq 0,$$ and because each product of two entries in \{0,1\} can be either 0 or 1, the determinant vanishes precisely when $$a_{11}a_{22} = a_{12}a_{21}.$$

Next, the probability that a product of two bits is 0 is $$\tfrac{3}{4}$$ (at least one factor is 0) and that it is 1 is $$\tfrac{1}{4}$$ (both are 1). Therefore, the probability that both products are 0 is $$\tfrac{3}{4}\cdot\tfrac{3}{4}=\tfrac{9}{16}$$, and that both are 1 is $$\tfrac{1}{4}\cdot\tfrac{1}{4}=\tfrac{1}{16}$$. Adding these gives $$P(\det(A)=0)=\tfrac{9}{16}+\tfrac{1}{16}=\tfrac{10}{16}.$$

Hence the probability that the matrix is invertible is $$1 - \tfrac{10}{16} = \tfrac{6}{16} = \tfrac{3}{8}.$$

The correct answer is Option (3): $$\frac{3}{8}$$.

The expression for $$C_{jk}$$ is the definition of a determinant if $$j=i$$, but here the indices are specific. Note that $$C = A^T \cdot \text{adj}(A)^T$$. However, a simpler property of cofactors is that $$A \cdot \text{adj}(A) = |A|I$$.

Looking at the sum $$C_{jk} = \sum a_{ik}A_{ik}$$, this represents the product of elements of a column and cofactors of a column.

$$|A| = (\log_5 128)(\log_4 25) - (\log_4 5)(\log_5 8)$$

Using base change: $$\log_b a = \frac{\ln a}{\ln b}$$

$$|A| = (\frac{7 \ln 2}{\ln 5})(\frac{2 \ln 5}{2 \ln 2}) - (\frac{\ln 5}{2 \ln 2})(\frac{3 \ln 2}{\ln 5})$$

$$|A| = 7 - \frac{3}{2} = \frac{11}{2} = 5.5$$

The matrix $$C$$ results in $$|C| = |A|^2$$ (due to the property of the product of $$A$$ and its cofactor matrix).

$$|C| = (\frac{11}{2})^2 = \frac{121}{4}$$

$$8|C| = 8 \times \frac{121}{4} = 2 \times 121 = \mathbf{242}$$

Correct Option: C

Let $$\Sigma(M)$$ denote the sum of all the elements of a matrix $$M$$.
We have $$A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ and the relation

$$A^{n} = A^{\,n-2} + A^{2} - I \quad\text{for}\; n \ge 3$$ $$-(1)$$

Applying $$\Sigma$$ on both sides of $$(1)$$ gives a relation for the sums:

$$\Sigma\!\left(A^{n}\right) = \Sigma\!\left(A^{\,n-2}\right) + \Sigma\!\left(A^{2}\right) - \Sigma(I)$$ $$-(2)$$

Because $$\Sigma$$ is linear (sum of elements of a sum is the sum of the individual sums), $$(2)$$ is valid for every $$n \ge 3$$.

First compute the required base values.

Case 1: $$A^{0}=I$$
$$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ so $$\Sigma(I)=1+1+1=3$$.

Case 2: $$A^{1}=A$$
Row sums: $$1,\;2,\;1 \;\Rightarrow\; \Sigma(A)=1+2+1=4$$.

Case 3: $$A^{2}$$

Compute $$A^{2}=A\! \cdot\! A$$:

$$A^{2}= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$
Row sums: $$1,\;2,\;2 \;\Rightarrow\; \Sigma(A^{2})=1+2+2=5$$.

Substituting $$\Sigma(A^{2})=5$$ and $$\Sigma(I)=3$$ into $$(2)$$ gives the recurrence for $$S_n=\Sigma(A^{n})$$:

$$S_n = S_{\,n-2} + 5 - 3 = S_{\,n-2} + 2 \quad\text{for}\; n \ge 3$$ $$-(3)$$

Now build the sequence.

Even powers
$$S_0 = 3$$ (already obtained)
Using $$(3)$$ repeatedly:
$$S_2 = S_0 + 2 = 3 + 2 = 5$$
$$S_4 = S_2 + 2 = 5 + 2 = 7$$
Continuing, every step of two in the index adds 2 to the sum, so

$$S_{2k} = 3 + 2k \quad\text{for}\; k \ge 0$$ $$-(4)$$

Odd powers
$$S_1 = 4$$ (already obtained)
Similarly,

$$S_3 = S_1 + 2 = 4 + 2 = 6$$
$$S_5 = S_3 + 2 = 6 + 2 = 8$$
Hence

$$S_{2k+1} = 4 + 2k \quad\text{for}\; k \ge 0$$ $$-(5)$$

We need $$S_{50}$$. Since $$50 = 2 \times 25$$ is even, use $$(4)$$ with $$k = 25$$:

$$S_{50} = 3 + 2 \times 25 = 3 + 50 = 53$$.

Therefore, the sum of all the elements of $$A^{50}$$ is $$53$$.

Option A is correct.

We need to find $$a_{23}$$ given three conditions on matrix A.

Write the conditions

Let A = $$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

From the given conditions:

$$A\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}0\\0\\1\end{bmatrix}$$ gives us the second column of A: $$a_{12} = 0, a_{22} = 0, a_{32} = 1$$

$$A\begin{bmatrix}4\\1\\3\end{bmatrix} = \begin{bmatrix}0\\1\\0\end{bmatrix}$$ gives:

$$4a_{11} + a_{12} + 3a_{13} = 0$$ ... (i)

$$4a_{21} + a_{22} + 3a_{23} = 1$$ ... (ii)

$$4a_{31} + a_{32} + 3a_{33} = 0$$ ... (iii)

$$A\begin{bmatrix}2\\1\\2\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$$ gives:

$$2a_{11} + a_{12} + 2a_{13} = 1$$ ... (iv)

$$2a_{21} + a_{22} + 2a_{23} = 0$$ ... (v)

$$2a_{31} + a_{32} + 2a_{33} = 0$$ ... (vi)

Find $$a_{23}$$

From (ii): $$4a_{21} + 0 + 3a_{23} = 1 \Rightarrow 4a_{21} + 3a_{23} = 1$$

From (v): $$2a_{21} + 0 + 2a_{23} = 0 \Rightarrow a_{21} = -a_{23}$$

Substituting in (ii): $$4(-a_{23}) + 3a_{23} = 1$$

$$-4a_{23} + 3a_{23} = 1$$

$$-a_{23} = 1$$

$$a_{23} = -1$$

The correct answer is Option 1: -1.

For a system of three linear equations to possess infinitely many solutions, the following two conditions must hold:
  • The determinant of the coefficient matrix must be zero (so its rank is < 3).
  • Every equation must be a linear combination of the others, i.e. the augmented matrix must have the same rank as the coefficient matrix.

Write the coefficient matrix $$A$$ and the constant column $$\mathbf{b}$$:

$$A = \begin{bmatrix}1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu\end{bmatrix},\qquad \mathbf{b} = \begin{bmatrix}2 \\ 5 \\ 33\end{bmatrix}.$$

Step 1 : Determinant of the coefficient matrix

The determinant is

$$\begin{vmatrix}1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu\end{vmatrix} = 1(\lambda\mu-5\!\cdot\!3)\;-\;2(2\mu-5\!\cdot\!14)\;+\;(-3)(2\!\cdot\!3-\lambda\!\cdot\!14).$$

Simplifying term by term:

$$\lambda\mu-15\;-\;2(2\mu-70)\;-\;3(6-14\lambda)$$

$$=\lambda\mu-15\;-\;4\mu+140\;-\,18+42\lambda$$

$$=\mu(\lambda-4)+42\lambda+107.$$

For infinitely many solutions, this determinant must vanish:

$$\mu(\lambda-4)+42\lambda+107=0\quad -(1).$$

Step 2 : Ensuring the third equation is a linear combination of the first two

Assume constants $$a$$ and $$b$$ exist such that

$$a(x+2y-3z)+b(2x+\lambda y+5z)=14x+3y+\mu z \quad\text{and}\quad a(2)+b(5)=33.$$(The left-hand side recreates the third equation’s coefficients and constant term.)

Matching the coefficients of $$x,\,y,\,z$$ and the constants gives

$$\begin{aligned} a+2b &= 14 \quad &(x\text{-coeff})\\ 2a+\lambda b &= 3 \quad &(y\text{-coeff})\\ -3a+5b &= \mu \quad &(z\text{-coeff})\\ 2a+5b &= 33 \quad &(\text{constant}) \end{aligned}$$

From $$a+2b=14$$ obtain

$$a = 14-2b \quad -(2).$$

Substitute $$a$$ from $$(2)$$ into the $$y$$-coefficient condition:

$$2(14-2b)+\lambda b = 3$$

$$28-4b+\lambda b = 3$$

$$(\lambda-4)b = -25$$

$$b = \frac{-25}{\lambda-4} \quad -(3).$$

Insert $$b$$ from $$(3)$$ into the constant condition $$2a+5b=33$$. First compute $$2a$$ using $$(2)$$:

$$2a = 2(14-2b) = 28-4b.$$

Hence

$$28-4b+5b = 33 \;\Longrightarrow\; 28 + b = 33$$

$$b = 5.$$

Equating this value of $$b$$ to the one in $$(3)$$ gives

$$5 = \frac{-25}{\lambda-4}$$

$$\lambda-4 = -5$$

$$\lambda = -1.$$

With $$\lambda=-1$$, use $$(3)$$ to confirm $$b=5$$ and $$(2)$$ to find $$a$$:

$$a = 14 - 2(5) = 4.$$

Finally, compute $$\mu$$ from the $$z$$-coefficient relation:

$$\mu = -3a + 5b = -3(4)+5(5) = -12+25 = 13.$$

Step 3 : Value of $$\lambda+\mu$$

$$\lambda+\mu = (-1)+13 = 12.$$

Therefore, $$\lambda+\mu = 12$$, which corresponds to Option C.

Since $$\alpha$$ satisfies $$x^{2}+x+1=0$$, we have
$$\alpha^{2}+\alpha+1=0$$ and hence $$\alpha^{3}=1,\;\alpha\neq 1.$$

The condition
$$[4\;a\;b]\, \begin{bmatrix} 1 & 16 & 13\\ -1& -1 & 2\\ -2&-14&-8 \end{bmatrix} =[0\;0\;0] $$ means the row-vector $$[4\;a\;b]$$ is in the left-null-space of the matrix. Multiplying out gives three linear equations:

$$4(1)+a(-1)+b(-2)=0\;\;\Longrightarrow\;\;4-a-2b=0\;\;\Longrightarrow\;\;a+2b=4\; -(1)$$
$$4(16)+a(-1)+b(-14)=0\;\;\Longrightarrow\;\;64-a-14b=0\;\;\Longrightarrow\;\;a+14b=64\; -(2)$$
$$4(13)+a(2)+b(-8)=0\;\;\Longrightarrow\;\;52+2a-8b=0\;\;\Longrightarrow\;\;a-4b=-26\; -(3)$$

Solving $$(1)$$ and $$(2)$$:
$$\bigl(a+14b\bigr)-\bigl(a+2b\bigr)=64-4\;\;\Longrightarrow\;\;12b=60\;\;\Longrightarrow\;\;b=5.$$

Substituting $$b=5$$ in $$(1)$$: $$a+2(5)=4\;\;\Longrightarrow\;\;a=-6.$$ (Equation $$(3)$$ is also satisfied, so the solution is consistent.)

Now evaluate each reciprocal power of $$\alpha$$ needed in the expression $$\frac{4}{\alpha^{4}}+\frac{m}{\alpha^{a}}+\frac{n}{\alpha^{5}}=3.$$ Because $$\alpha^{3}=1,$$ we can reduce every exponent modulo $$3$$:

$$\frac{1}{\alpha^{4}}=\frac{1}{\alpha^{3}\alpha}=\frac{1}{\alpha}= \alpha^{2},$$
$$\frac{1}{\alpha^{a}}=\frac{1}{\alpha^{-6}}=\alpha^{6}= (\alpha^{3})^{2}=1,$$
$$\frac{1}{\alpha^{5}}=\frac{1}{\alpha^{3}\alpha^{2}}=\frac{1}{\alpha^{2}}=\alpha.$$

Substituting these values converts the given relation to a polynomial in $$\alpha$$:

$$4\alpha^{2}+m\cdot 1+n\alpha = 3 \;\;\Longrightarrow\;\; 4\alpha^{2}+n\alpha+(m-3)=0.$$

The above equation must hold for both roots of $$x^{2}+x+1=0.$$ A polynomial of degree $$2$$ that vanishes at both roots of another irreducible quadratic is necessarily a scalar multiple of that quadratic. Hence

$$4\alpha^{2}+n\alpha+(m-3)=k\bigl(\alpha^{2}+\alpha+1\bigr).$$

Comparing coefficients gives the common scalar $$k=4$$ and

$$n = k = 4, \qquad m-3 = k = 4 \;\;\Longrightarrow\;\; m = 7.$$

Therefore $$m+n = 7+4 = 11.$$

Option B (11)

The given matrix equation is $$A^2(A-2I)-4(A-I)=O$$.

Expand the left-hand side:
$$A^2(A-2I)-4(A-I)=A^3-2A^2-4A+4I=O.$$

Hence $$A^3-2A^2-4A+4I=O \quad\Longrightarrow\quad A^3=2A^2+4A-4I \; -(1).$$

Step 1: Find $$A^4$$.
Multiply both sides of $$(1)$$ by $$A$$ on the left:
$$A^4=A\bigl(2A^2+4A-4I\bigr)=2A^3+4A^2-4A.$$

Replace $$A^3$$ using $$(1)$$ again:
$$A^4=2(2A^2+4A-4I)+4A^2-4A\\ \phantom{A^4}=4A^2+8A-8I+4A^2-4A\\ \phantom{A^4}=8A^2+4A-8I \; -(2).$$

Step 2: Find $$A^5$$.
Multiply $$(2)$$ by $$A$$ on the left:
$$A^5=A\bigl(8A^2+4A-8I\bigr)=8A^3+4A^2-8A.$$

Again substitute $$A^3$$ from $$(1)$$:
$$A^5=8(2A^2+4A-4I)+4A^2-8A\\ \phantom{A^5}=16A^2+32A-32I+4A^2-8A\\ \phantom{A^5}=20A^2+24A-32I.$$

Thus $$A^5=\alpha A^2+\beta A+\gamma I$$ with
$$\alpha=20,\qquad \beta=24,\qquad \gamma=-32.$$

Therefore $$\alpha+\beta+\gamma=20+24-32=12.$$

Hence the required value is $$12$$, which corresponds to Option A.

$$A$$ is a $$3 \times 3$$ matrix with $$\det(A) = -2$$. We need $$\det(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$$ with $$m > n$$.

For an $$n \times n$$ matrix (here $$n = 3$$):

$$\det(kA) = k^n \det(A)$$, $$\det(\text{adj}(A)) = (\det(A))^{n-1}$$, $$\text{adj}(kA) = k^{n-1} \text{adj}(A)$$

$$\det(3A) = 3^3 \det(A) = 27 \times (-2) = -54$$

$$\det(\text{adj}(3A)) = (\det(3A))^{3-1} = (-54)^2 = 2916$$

$$\det(-6 \cdot \text{adj}(3A)) = (-6)^3 \cdot \det(\text{adj}(3A)) = -216 \times 2916$$

$$= -216 \times 2916 = -629856$$

$$= (\det(-6 \cdot \text{adj}(3A)))^2 = (-629856)^2 = 629856^2$$

$$= 3^3 \times 629856^2 = 27 \times 629856^2$$

Now factorise: $$629856 = 216 \times 2916 = 6^3 \times 54^2 = (2 \cdot 3)^3 \times (2 \cdot 3^3)^2 = 2^3 \cdot 3^3 \cdot 2^2 \cdot 3^6 = 2^5 \cdot 3^9$$

$$629856^2 = 2^{10} \cdot 3^{18}$$

$$27 \times 629856^2 = 3^3 \times 2^{10} \times 3^{18} = 2^{10} \times 3^{21}$$

So $$2^{m+n} \cdot 3^{mn} = 2^{10} \cdot 3^{21}$$ with $$m > n$$.

$$m + n = 10$$ and $$mn = 21$$.

Solving: $$m$$ and $$n$$ are roots of $$t^2 - 10t + 21 = 0$$, giving $$t = 7$$ or $$t = 3$$.

Since $$m > n$$: $$m = 7, n = 3$$.

$$4m + 2n = 28 + 6 = 34$$.

The answer is 34.

The determinant of the given matrix $$A=\begin{bmatrix}\lambda&2&3\\4&5&6\\7&-1&2\end{bmatrix}$$ is

$$|A|=\lambda\bigl(5\cdot2-6\cdot(-1)\bigr)-2\bigl(4\cdot2-6\cdot7\bigr)+3\bigl(4\cdot(-1)-5\cdot7\bigr)$$
$$\;\;=\lambda(10+6)-2(8-42)+3(-4-35)=16\lambda-49.$$

Given $$|A|=-1,$$ we get $$16\lambda-49=-1\; \Longrightarrow\; \lambda=3.$$

Case 1: Simplifying $$A\cdot\text{adj}(A^{2})$$

For any non-singular matrix, $$\text{adj}(A)=|A|\,A^{-1}.$$ Hence

$$\text{adj}(A^{2})=|A^{2}|(A^{2})^{-1}=|A|^{2}A^{-2}.$$(Because $$|A^{2}|=|A|^{2}.$$)

Therefore

$$A\cdot\text{adj}(A^{2})=A\,\bigl(|A|^{2}A^{-2}\bigr)=|A|^{2}A^{-1}.$$

With $$|A|=-1,\quad |A|^{2}=1,$$ so

$$A\cdot\text{adj}(A^{2})=A^{-1}.$$ Denote $$M=A^{-1}.$$

Case 2: Finding the matrix $$B$$

First observe

$$\text{adj}(M)=\text{adj}(A^{-1})=|A^{-1}|\,A=(|A|)^{-1}A=(-1)^{-1}A=-A.$$ (The determinant $$|A^{-1}|=(|A|)^{-1}=-1.$)

By definition, $$B=\bigl($$\text{adj}$$(M)\bigr)^{-1}=(-A)^{-1}=-A^{-1}.$$

Case 3: Evaluating $$|($$\lambda$$ B+I)|$$

Put $$$$\lambda$$=3$$ and $$B=-A^{-1}$$:

$$$$\lambda$$ B+I=3(-A^{-1})+I=I-3A^{-1}.$$

Factor out $$A^{-1}$$: $$I-3A^{-1}=A^{-1}\,(A-3I).$$

Hence, for a $$3$$\times$$3$$ matrix,

$$|\,I-3A^{-1}\,|=|A^{-1}|\,|A-3I|.$$

We already know $$|A^{-1}|=(|A|)^{-1}=-1.$$ So we only need $$|A-3I|$$.

Compute $$A-3I$$ when $$$$\lambda$$=3$$:
$$A-3I=$$\begin{bmatrix}$$0&2&3\\4&2&6\\7&-1&-1\end{bmatrix}.$$ Its determinant is

$$0\bigl(2$$\cdot$$(-1)-6$$\cdot$$(-1)\bigr)-2\bigl(4$$\cdot$$(-1)-6$$\cdot$$7\bigr)+3\bigl(4$$\cdot$$(-1)-2$$\cdot$$7\bigr)=38.$$

Therefore

$$|\,$$\lambda$$ B+I\,|=|\,I-3A^{-1}\,|=(-1)$$\times$$38=-38.$$

The numerical value (modulus) of the determinant is $$38.$$

Final answer = 38

Write $$A$$ in compact form by denoting $$c=\cos\theta$$ and $$s=\sin\theta$$:

$$A=\begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix}$$

First compute $$A^T$$ and $$A^2$$.

Transpose:
$$A^T=\begin{bmatrix} c & 0 & s \\ 0 & 1 & 0 \\ -s & 0 & c \end{bmatrix}$$

Square of $$A$$:
$$A^2=\begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix} \begin{bmatrix} c & 0 & -s \\ 0 & 1 & 0 \\ s & 0 & c \end{bmatrix} =\begin{bmatrix} c^2-s^2 & 0 & -2cs \\ 0 & 1 & 0 \\ 2cs & 0 & c^2-s^2 \end{bmatrix}$$

Using the double-angle identities, rewrite

$$A^2=\begin{bmatrix} \cos 2\theta & 0 & -\sin 2\theta \\ 0 & 1 & 0 \\ \sin 2\theta & 0 & \cos 2\theta \end{bmatrix}$$

The condition $$A^2=A^T$$ gives three independent equations:

$$\cos 2\theta = \cos\theta$$
$$-\sin 2\theta = \sin\theta$$
$$\sin 2\theta = -\sin\theta$$

Because $$\theta\in(0,\pi)$$, $$\sin\theta\neq0$$. Divide the second equation by $$\sin\theta$$:

$$-2\cos\theta = 1 \;\;\Longrightarrow\;\; \cos\theta = -\frac12$$

Thus $$\theta=\frac{2\pi}{3}$$, giving

$$c=-\frac12,\qquad s=\frac{\sqrt3}{2}$$

Next define the required polynomial in $$A$$:

$$E = (A+I)^3 + (A-I)^3 - 6A$$

Expand the two cubes separately:

$$(A+I)^3 = A^3 + 3A^2 + 3A + I$$
$$(A-I)^3 = A^3 - 3A^2 + 3A - I$$

Add them and subtract $$6A$$:

$$E = \big(A^3 + 3A^2 + 3A + I\big) + \big(A^3 - 3A^2 + 3A - I\big) - 6A$$
$$\;\; = 2A^3 + 6A - 6A = 2A^3$$

Therefore $$E=2A^3$$ and the trace of $$E$$ is simply twice the trace of $$A^3$$.

To find $$\operatorname{tr}(A^3)$$, use the eigenvalues of $$A$$. A$$ is a rotation matrix about the $$y$$-axis through $$$$\theta=\frac{2\pi}{3}$$$$, so its eigenvalues are

$$$$\lambda_1$$ = 1,\qquad $$\lambda_2$$ = e^{i$$\theta$$},\qquad $$\lambda_3$$ = e^{-i$$\theta$$}$$

Hence

$$$$\lambda_2^3$$ = e^{3i$$\theta$$}=e^{i2$$\pi$$}=1,\qquad $$\lambda_3^3$$ = e^{-3i$$\theta$$}=1$$

Thus

$$\operatorname{tr}(A^3)=$$\lambda_1^3+\lambda_2^3+\lambda_3^3$$ = 1+1+1 = 3$$

Finally,

$$\operatorname{tr}(E)=2\,\operatorname{tr}(A^3)=2$$\times$$3=6$$

Hence the sum of the diagonal elements of $$(A + I)^3 + (A - I)^3 - 6A$$ equals $$6$$.

We need to find $$\alpha$$ where $$n(S_1 \cup S_2 \cup S_3) = 125\alpha$$. Since $$S = \{-3, -2, -1, 1, 2\}$$ contains 5 elements and all matrices are $$3 \times 3$$ with entries from $$S$$, we proceed to count each set.

For $$S_1$$, the symmetric matrices $$A = A^T$$ satisfy $$a_{ij} = a_{ji}$$ so we may choose freely the six entries $$a_{11},\,a_{12},\,a_{13},\,a_{22},\,a_{23},\,a_{33}$$ each having 5 possible values. This gives $$|S_1| = 5^6 = 15625$$.

In the case of skew‐symmetric matrices $$S_2$$ with $$A = -A^T$$ the diagonal entries must satisfy $$a_{ii} = 0$$, but since $$0\notin S$$ no such matrices exist and hence $$|S_2| = 0$$.

Considering $$S_3$$ of trace‐zero matrices where $$a_{11} + a_{22} + a_{33} = 0$$, the six off‐diagonal entries remain free with $$5^6$$ choices. We then count the ordered triples $$(a_{11},a_{22},a_{33}) \in S^3$$ summing to zero. Among the $$5^3 = 125$$ possible triples only $$(-3,1,2)$$ in 6 permutations, $$(-2,1,1)$$ in 3 permutations, and $$(-1,-1,2)$$ in 3 permutations work, giving 12 solutions. Therefore $$|S_3| = 12 \times 5^6 = 187500$$.

Since $$S_2$$ is empty we have $$|S_1 \cap S_2| = 0$$ and $$|S_2 \cap S_3| = 0$$. For $$S_1 \cap S_3$$, symmetric matrices with trace zero allow the three off‐diagonal entries $$a_{12}, a_{13}, a_{23}$$ to be chosen arbitrarily in $$5^3 = 125$$ ways, while the diagonal triple must sum to zero in 12 ways, yielding $$|S_1 \cap S_3| = 12 \times 125 = 1500$$. Moreover, the triple intersection $$|S_1 \cap S_2 \cap S_3|$$ is also 0.

By inclusion-exclusion we obtain $$|S_1 \cup S_2 \cup S_3| = 15625 + 0 + 187500 - 0 - 1500 - 0 + 0 = 201625$$.

Substituting into $$125\alpha = |S_1 \cup S_2 \cup S_3|$$ gives $$125\alpha = 201625$$, which yields $$\alpha = \frac{201625}{125} = 1613$$.

Option X: The answer is 1613.

Since $$X^TAX = 0$$ for all nonzero $$3 \times 1$$ matrices $$X$$, the matrix $$A$$ must be skew-symmetric: $$A^T = -A$$. A $$3 \times 3$$ skew-symmetric matrix has the form $$A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$$.

From $$A\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}1\\4\\-5\end{pmatrix}$$, we get $$a + b = 1$$, $$-a + c = 4$$, and $$-b - c = -5$$. From $$A\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}0\\4\\-8\end{pmatrix}$$, the first component gives $$2a + b = 0$$, so $$a = -1$$, $$b = 2$$, $$c = 3$$.

Now $$A + I = \begin{pmatrix} 1 & -1 & 2 \\ 1 & 1 & 3 \\ -2 & -3 & 1 \end{pmatrix}$$ and $$2(A+I) = \begin{pmatrix} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -4 & -6 & 2 \end{pmatrix}$$. Computing $$\det(2(A+I)) = 2(4+36) + 2(4+24) + 4(-12+8) = 80 + 56 - 16 = 120$$. For a $$3 \times 3$$ matrix, $$\det(\text{adj}(B)) = (\det B)^{n-1} = 120^2 = 14400 = 2^6 \cdot 3^2 \cdot 5^2$$.

Thus $$\alpha = 6$$, $$\beta = 2$$, $$\gamma = 2$$, and $$\alpha^2 + \beta^2 + \gamma^2 = 36 + 4 + 4 = \boxed{44}$$.

We are given $$A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}$$ and $$S = \left\{m \in \mathbb{Z} : A^{m^2} + A^m = 3I - A^{-6}\right\}$$.

The characteristic equation of A: $$\det(A - \lambda I) = (2-\lambda)(0-\lambda) - (-1)(1) = \lambda^2 - 2\lambda + 1 = (\lambda - 1)^2 = 0$$.

By the Cayley-Hamilton theorem: $$(A - I)^2 = 0$$, so $$A^2 - 2A + I = 0$$, meaning $$A^2 = 2A - I$$.

Since $$(A - I)^2 = 0$$, let $$N = A - I$$, so $$N^2 = 0$$. Then $$A = I + N$$.

Let us verify: $$A^2 = -I + 2A = 2A - I$$. This matches $$A^2 = 2A - I$$. Good.

So $$A^n = (1-n)I + nA$$ for all integers $$n$$.

$$A^{m^2} + A^m = 3I - A^{-6}$$

Since $$I$$ and $$A$$ are linearly independent (as $$A \ne kI$$):

Coefficient of $$A$$: $$m^2 + m = 6$$, so $$m^2 + m - 6 = 0$$, giving $$(m+3)(m-2) = 0$$, so $$m = -3$$ or $$m = 2$$.

Coefficient of $$I$$: $$2 - m^2 - m = -4$$, so $$m^2 + m = 6$$. This gives the same equation.

Both conditions are identical, so $$m \in \{-3, 2\}$$ and $$n(S) = 2$$.

The answer is $$\boxed{2}$$.

The two conditions on the integer matrix $$Q$$ are

1. $$Q^{-1}=Q^{T}$$    ⇒  $$Q$$ is an orthogonal matrix with integer entries.
2. $$PQ=QP$$ with $$P=\begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}$$.

Step 1: Characterise all $$3\times3$$ integer orthogonal matrices.
If $$Q$$ is orthogonal and its entries are integers, each row (and each column) is a unit vector whose components are integers. The only integer vectors of length 1 are the six vectors$$(\pm1,0,0),\;(0,\pm1,0),\;(0,0,\pm1).$$Hence every row and every column of $$Q$$ contains exactly one entry $$\pm1$$ and the rest are zeros. Such matrices are called signed permutation matrices.

The total number of signed permutation matrices of order 3 is $$3!\times2^{3}=6\times8=48$$ (choose a permutation of the columns and then choose the sign of the non-zero entry in each column).

Step 2: Impose the commutation condition $$PQ=QP$$.
Write $$Q=(q_{ij})$$. From $$PQ=QP$$ we get, for every pair $$(i,j),$$ $$P_{ii}\,q_{ij}=q_{ij}\,P_{jj}.$$
Because $$P=\text{diag}(2,2,3),$$ this gives $$(P_{ii}-P_{jj})\,q_{ij}=0\quad\forall\,i,j.$$ Whenever $$P_{ii}\neq P_{jj},$$ the factor $$P_{ii}-P_{jj}$$ is non-zero, forcing $$q_{ij}=0.$

Thus every entry mixing the first two coordinates with the third coordinate must vanish: $$q_{13}=q_{23}=q_{31}=q_{32}=0.$$ Therefore $$Q$$ has the block-diagonal form $$Q=$$\begin{pmatrix}$$ \ast & \ast & 0\\ \ast & \ast & 0\\ 0 & 0 & \ast \end{pmatrix},$$ where the asterisks are the remaining possible $$$$\pm$$1$$ or $$0$$ entries.

Step 3: Count the admissible signed permutation matrices that satisfy the block structure.
• The lower-right $$1$$\times$$1$$ block must be $$$$\pm$$1$$ (two choices).
• The upper-left $$2$$\times$$2$$ block must itself be a signed permutation matrix (to keep orthogonality and to make each column/row have exactly one non-zero entry).
  - There are $$2! = 2$$ ways to permute the two columns.
  - Independently, each of the two non-zero entries can be $$+1$$ or $$-1$$, giving $$2^{2}=4$$ sign choices.

Hence the number of possible $$2$$\times$$2$$ blocks is $$2$$\times$$4=8$$.
Multiplying by the two choices in the $$1$$\times$$1$$ block, we obtain the final count $$8$$\times$$2 = 16.$$

Step 4: Conclusion.
Exactly $$16$$ integer matrices $$Q$$ satisfy both $$Q^{-1}=Q^{T}$$ and $$PQ=QP$$.

Option C which is: $$16$$

Property: For any non-singular matrix $$X$$, $$adj(X) = |X|X^{-1}$$.

Simplify $$adj(A^{-1}) = |A^{-1}|(A^{-1})^{-1} = \frac{1}{|A|}A$$.

Let $$K = adj(A^{-1}) + adj(B^{-1}) = \frac{A}{|A|} + \frac{B}{|B|}$$.

The expression is $$M = A \cdot K^{-1} \cdot B$$. We need $$M^{-1}$$.

$$M^{-1} = (A \cdot K^{-1} \cdot B)^{-1} = B^{-1} \cdot K \cdot A^{-1}$$

 Substitute $$K$$ back:

$$M^{-1} = B^{-1} \left( \frac{A}{|A|} + \frac{B}{|B|} \right) A^{-1} = \frac{B^{-1}AA^{-1}}{|A|} + \frac{B^{-1}BA^{-1}}{|B|} = \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$$

 Multiply and divide by $$|A||B|$$ to match options:

$$M^{-1} = \frac{|B|B^{-1} + |A|A^{-1}}{|A||B|} = \frac{adj(B) + adj(A)}{|AB|}$$

Option D is correct.

The matrix $$A = [a_{ij}]$$ is defined by $$a_{ij} = (\sqrt{2})^{i+j}$$.

$$ A = \begin{bmatrix} 2 & 2\sqrt{2} & 4 \\ 2\sqrt{2} & 4 & 4\sqrt{2} \\ 4 & 4\sqrt{2} & 8 \end{bmatrix} $$

We seek the sum of all elements in the third row of $$A^2$$.

Notice that the sum of the entries in the third row of $$A^2$$ can be written as the dot product of the third row of $$A$$ with the vector of column sums of $$A$$ (i.e., $$A^2 \cdot \mathbf{1}$$ gives row sums of $$A^2$$, which equals $$A \cdot (A\mathbf{1})$$).

The sums of the columns of $$A$$ are

Column 1: $$2 + 2\sqrt{2} + 4 = 6 + 2\sqrt{2}$$

Column 2: $$2\sqrt{2} + 4 + 4\sqrt{2} = 4 + 6\sqrt{2}$$

Column 3: $$4 + 4\sqrt{2} + 8 = 12 + 4\sqrt{2}$$

Hence the sum of the third row of $$A^2$$ is

$$=4(6+2\sqrt{2}) + 4\sqrt{2}(4+6\sqrt{2}) + 8(12+4\sqrt{2})$$

$$=24 + 8\sqrt{2} + 16\sqrt{2} + 48 + 96 + 32\sqrt{2}$$

$$=(24 + 48 + 96) + (8 + 16 + 32)\sqrt{2}$$

$$=168 + 56\sqrt{2}$$

So $$\alpha = 168$$ and $$\beta = 56$$.

$$\alpha + \beta = 168 + 56 = 224$$

The correct answer is Option 2: 224.

We are given $$A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix}$$ with $$\alpha \gt 0$$, $$\det(A) = 0$$, and $$\alpha + \beta = 1$$.

From $$\det(A) = 0$$: $$\alpha\beta + 6 = 0$$, so $$\alpha\beta = -6$$.

From $$\alpha + \beta = 1$$: $$\beta = 1 - \alpha$$. Substituting: $$\alpha(1 - \alpha) = -6$$, giving $$\alpha - \alpha^2 = -6$$, so $$\alpha^2 - \alpha - 6 = 0$$.

Factoring: $$(\alpha - 3)(\alpha + 2) = 0$$. Since $$\alpha \gt 0$$, we get $$\alpha = 3$$ and $$\beta = -2$$.

So $$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}$$ and $$I + A = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix}$$.

Let $$B = I + A$$. We compute $$B^2$$: $$B^2 = \begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix}\begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} = \begin{bmatrix} 16-6 & -4+1 \\ 24-6 & -6+1 \end{bmatrix} = \begin{bmatrix} 10 & -3 \\ 18 & -5 \end{bmatrix}$$.

Note that $$\text{tr}(B) = 3$$ and $$\det(B) = -4 + 6 = 2$$. By Cayley-Hamilton, $$B^2 = 3B - 2I$$, i.e., $$B^2 - 3B + 2I = 0$$.

We can express higher powers using the recurrence $$B^n = 3B^{n-1} - 2B^{n-2}$$. The characteristic equation $$\lambda^2 - 3\lambda + 2 = 0$$ has roots $$\lambda = 1$$ and $$\lambda = 2$$.

So $$B^n = \alpha_0 \cdot 1^n \cdot I + \alpha_1$$ ... Let us write $$B^n = c_1 \cdot 2^n I + c_2 \cdot 1^n I$$ — actually since the eigenvalues are 1 and 2, we write $$B^n = (2^n - 1)(B - I) + (2 \cdot 1^n - 1 \cdot 2^n)(- I) + ...$$ Let us use the direct formula.

Since $$B$$ has eigenvalues 1 and 2, we can write $$B = P \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} P^{-1}$$, so $$B^n = P \begin{bmatrix} 1 & 0 \\ 0 & 2^n \end{bmatrix} P^{-1}$$.

Using the formula $$B^n = \frac{2^n(B - I) - 1^n(B - 2I)}{2 - 1} = 2^n(B - I) - (B - 2I) = (2^n - 1)B - (2^n - 2)I$$.

For $$n = 8$$: $$B^8 = (256 - 1)B - (256 - 2)I = 255B - 254I$$.

$$B^8 = 255\begin{bmatrix} 4 & -1 \\ 6 & -1 \end{bmatrix} - 254\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1020 - 254 & -255 \\ 1530 & -255 - 254 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\ 1530 & -509 \end{bmatrix}$$.

Hence, the correct answer is Option D.

The condition $$QR = RP$$ will be used to connect the unknown entries of $$Q$$ with the (non-zero) entries of $$R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$.

Compute both products:

$$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} xa + yc & xb + yd \\ za + 4c & zb + 4d \end{pmatrix}$$

$$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \!\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$$

Equating corresponding entries of $$QR$$ and $$RP$$ yields four linear equations:

$$\begin{aligned} xa + yc &= 2a \quad &-(1)\\ xb + yd &= 3b \quad &-(2)\\ za + 4c &= 2c \quad &-(3)\\ zb + 4d &= 3d \quad &-(4) \end{aligned}$$

From $$(3)$$ and $$(4)$$ obtain $$z$$ in two ways:

$$z = \frac{-2c}{a} \quad -(5), \qquad z = \frac{-d}{b} \quad -(6)$$

Equating $$(5)$$ and $$(6)$$ gives the relation $$2bc = ad \quad -(7)$$.

Now express $$x$$ from $$(1)$$ and $$(2)$$:

$$x = 2 - \frac{c}{a}\,y \quad -(8), \qquad x = 3 - \frac{d}{b}\,y \quad -(9)$$

Equating $$(8)$$ and $$(9)$$ gives $$y\Bigl(\frac{d}{b} - \frac{c}{a}\Bigr) = 1 \quad -(10)$$

Use $$(7)$$ to rewrite $$\frac{d}{b} = \frac{2c}{a}$$, so the bracket in $$(10)$$ becomes $$\frac{2c}{a} - \frac{c}{a} = \frac{c}{a}.$$ Hence $$y \cdot \frac{c}{a} = 1 \;\;\Longrightarrow\;\; y = \frac{a}{c} \quad -(11)$$

Substitute $$(11)$$ into $$(8)$$: $$x = 2 - \frac{c}{a}\cdot\frac{a}{c} = 2 - 1 = 1 \quad -(12)$$

Finally, from $$(5)$$ and $$(11)$$ obtain $$z = \frac{-2c}{a}, \qquad yz = \frac{a}{c}\cdot\frac{-2c}{a} = -2 \quad -(13)$$

Thus the matrix $$Q$$ is $$Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}, \quad y\neq 0,\; z\neq 0,\; yz=-2.$$ (The specific non-zero values of $$a,b,c,d$$ simply scale $$y$$ and $$z$$ while keeping $$yz=-2$$.)

Determinant of $$Q - 2I$$
$$Q-2I = \begin{pmatrix} 1-2 & y \\ z & 4-2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$$ $$\det(Q-2I) = (-1)(2) - yz = -2 - (-2) = 0.$$ Statement A is true.

Determinant of $$Q - 6I$$
$$Q-6I = \begin{pmatrix} 1-6 & y \\ z & 4-6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$$ $$\det(Q-6I) = (-5)(-2) - yz = 10 - (-2) = 12.$$ Statement B is true.

Determinant of $$Q - 3I$$
$$Q-3I = \begin{pmatrix} 1-3 & y \\ z & 4-3 \end{pmatrix} = \begin{pmatrix} -2 & y \\ z & 1 \end{pmatrix}$$ $$\det(Q-3I) = (-2)(1) - yz = -2 - (-2) = 0\neq 15.$$ Statement C is false.

Product $$yz$$
From $$(13)$$, $$yz=-2\neq 2,$$ so Statement D is false.

Hence the correct statements are:
Option A  and  Option B.

The quadratic $$x^{2}+x-1=0$$ has distinct roots $$\alpha,\,\beta$$ satisfying the Vieta relations $$\alpha+\beta=-1$$ and $$\alpha\beta=-1$$. Consequently $$1+\alpha+\beta=0$$, a fact that will be used repeatedly.

Let a row be $$(t_{1},t_{2},t_{3})$$ with each $$t_{k}\in T=\{1,\alpha,\beta\}$$ and $$t_{1}+t_{2}+t_{3}=0$$. Write the counts of $$1,\alpha,\beta$$ in the row as $$(n_{1},n_{\alpha},n_{\beta})$$. Then $$n_{1}+n_{\alpha}+n_{\beta}=3$$ and $$n_{1}-n_{\beta}+\alpha\!\left(n_{\alpha}-n_{\beta}\right)=0.$$ Since $$1,\alpha$$ are linearly independent over $$\mathbb{Q}$$, we must have $$n_{1}=n_{\alpha}=n_{\beta}=1.$$ Thus every row is a permutation of $$(1,\alpha,\beta)$$, and the same holds for every column.

A $$3\times3$$ array in which each row and column contains every symbol exactly once is a Latin square of order $$3$$. For a fixed first row there are exactly $$2$$ such squares; there are $$3!=6$$ possible first rows, giving $$6\times2=12$$ matrices.

Therefore $$P\rightarrow 12\;.$$

Because the matrix is symmetric, row sums equal column sums, so every row also sums to $$0$$ and hence is a permutation of $$(1,\alpha,\beta)$$. Let the diagonal be $$(d_{1},d_{2},d_{3})$$. If any two diagonal entries were equal, the corresponding column would repeat that entry, violating the “one-of-each’’ rule; therefore $$d_{1},d_{2},d_{3}$$ are all distinct. There are $$3!=6$$ ways to choose this ordered diagonal.

Once the diagonal is fixed, the $$(i,j)$$-entry for $$i\lt j$$ must be the unique element of $$T$$ different from $$d_{i}$$ and $$d_{j}$$, and symmetry forces the $$(j,i)$$-entry to be the same. Hence the rest of the matrix is determined uniquely.

Thus there are $$6$$ symmetric matrices, so $$Q\rightarrow 6\;.$$

A $$3\times3$$ skew-symmetric matrix is of the form $$M=\begin{pmatrix}0&b&c\\-b&0&d\\-c&-d&0\end{pmatrix},$$ where $$b,c,d\in T$$ (because the entries below the diagonal are required to lie in $$T$$). The given system is $$M\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}b\\0\\-d\end{pmatrix}.$$

The second column of $$M$$ is $$\bigl[b,\,0,\,-d\bigr]^{\!T}$$, exactly the required right-hand side. Since the rank of a $$3\times3$$ skew-symmetric matrix is $$0$$ or $$2$$ (it is never full), this right-hand side is always in the column space, so the system is consistent. Because the rank is $$2$$, the solution space has dimension $$3-2=1$$, containing infinitely many vectors.

Hence the set in question is infinite, so $$R\rightarrow \text{Infinite}\;.$$

Each row sums to $$0$$, so (as in Case P) every row is a permutation of $$(1,\alpha,\beta)$$. Thus every row lies in the plane $$x+y+z=0$$, a $$2$$-dimensional subspace of $$\mathbb{R}^{3}$$. Three vectors confined to a plane are necessarily linearly dependent, so the matrix is singular and its determinant is $$0$$.

Therefore $$S\rightarrow 0\;.$$

Collecting the results:
(P) $$\to$$ 12 (Option 2), (Q) $$\to$$ 6 (Option 4), (R) $$\to$$ Infinite (Option 3), (S) $$\to$$ 0 (Option 5)

The option that matches this combination is
Option C: (P) → (2), (Q) → (4), (R) → (3), (S) → (5).

Given $$|A| = 3$$ and $$|B| = 2$$ for 3×3 matrices, we want to find $$|A^T A(\text{adj}(2A))^{-1}(\text{adj}(4B))(\text{adj}(AB))^{-1}AA^T|$$.

Some key properties for $$n \times n$$ matrices (with $$n = 3$$) are: $$|A^T| = |A|$$, $$|\text{adj}(M)| = |M|^{n-1} = |M|^2$$, $$|kM| = k^n|M| = k^3|M|$$, and $$|\text{adj}(kM)| = |kM|^2 = k^6|M|^2$$.

From these properties, one has $$|A^T| = 3$$ and $$|A| = 3$$. Moreover, $$|(\text{adj}(2A))^{-1}| = \frac{1}{|\text{adj}(2A)|} = \frac{1}{|2A|^2} = \frac{1}{(8 \times 3)^2} = \frac{1}{576}$$, while $$|\text{adj}(4B)| = |4B|^2 = (64 \times 2)^2 = 128^2 = 16384$$, and $$|(\text{adj}(AB))^{-1}| = \frac{1}{|AB|^2} = \frac{1}{(3 \times 2)^2} = \frac{1}{36}$$. Collecting these determinants gives

One can verify that $$81/20736 = 81/(81 \times 256) = 1/256$$, so $$16384/256 = 64$$. The correct answer is Option (4): 64.

We have $$A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$$ and $$A^3 = 4A^2 - A - 21I$$, which implies $$A^3 - 4A^2 + A + 21I = 0$$, so by the Cayley-Hamilton theorem $$A$$ satisfies its own characteristic equation.

$$p(\lambda) = \lambda^3 - 4\lambda^2 + \lambda + 21 = 0$$

The trace of $$A$$ is $$2 + 3 + b = 5 + b$$, and since the coefficient of $$\lambda^2$$ in the characteristic polynomial is the negative of the trace, we have $$-(5 + b) = -4$$ which gives $$5 + b = 4$$ and hence $$b = -1$$.

The sum of the cofactors of the diagonal entries equals the coefficient of $$\lambda$$ in the characteristic polynomial. The cofactor of $$a_{11}$$ is $$3b - 5 = -3 - 5 = -8$$, of $$a_{22}$$ is $$2b - 0 = -2$$, and of $$a_{33}$$ is $$6 - a$$, so their sum is $$-8 - 2 + 6 - a = -4 - a$$ which must equal $$1$$. Therefore $$-4 - a = 1$$ and hence $$a = -5$$.

To verify, the determinant of $$A$$, which equals the negative of the constant term of the characteristic polynomial, is $$2(3b - 5) - a(b) + 0 = 2(-8) -(-5)(-1) = -16 - 5 = -21$$. Since the constant term is $$-\det(A) = 21$$, the calculation is consistent.

Finally, $$2a + 3b = 2(-5) + 3(-1) = -10 - 3 = -13$$, so the correct answer is Option B: $$-13$$.

Given $$x\sin\theta = y\sin(\theta + 2\pi/3) = z\sin(\theta + 4\pi/3) \neq 0$$.

Let this common value be $$k$$. Then $$x = k/\sin\theta$$, $$y = k/\sin(\theta+2\pi/3)$$, $$z = k/\sin(\theta+4\pi/3)$$.

Trace(R) = $$x + y + z = k\left[\frac{1}{\sin\theta} + \frac{1}{\sin(\theta+2\pi/3)} + \frac{1}{\sin(\theta+4\pi/3)}\right]$$.

We know that $$\frac{1}{\sin\theta} + \frac{1}{\sin(\theta+2\pi/3)} + \frac{1}{\sin(\theta+4\pi/3)}$$ is generally NOT zero.

For example, at $$\theta = \pi/2$$: $$1 + 1/\sin(7\pi/6) + 1/\sin(11\pi/6) = 1 + (-2) + (-2) = -3 \neq 0$$.

So statement (I) is NOT always true.

For statement (II): adj(adj(R)) for a diagonal matrix has entries related to the (n-1)th powers of cofactors. For 3×3: adj(adj(R)) = det(R) · R (when R is invertible). Trace(adj(adj(R))) = det(R) · Trace(R).

If Trace(adj(adj(R))) = 0, then either det(R) = 0 or Trace(R) = 0. This doesn't necessarily mean R has exactly one non-zero entry. So (II) is also not necessarily true.

The answer is Option (3): Neither (I) nor (II) is true.

$$A$$ is a square matrix with $$AA^T = I$$, i.e., $$A$$ is orthogonal, so $$A^T = A^{-1}$$.

Find $$\frac{1}{2}A\left[(A+A^T)^2 + (A-A^T)^2\right]$$.

Expand the squares.

$$(A+A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2$$

$$(A-A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2$$

Add the two expressions.

$$(A+A^T)^2 + (A-A^T)^2 = 2A^2 + 2(A^T)^2$$

Multiply by $$\frac{1}{2}A$$.

$$\frac{1}{2}A \cdot [2A^2 + 2(A^T)^2] = A \cdot [A^2 + (A^T)^2] = A^3 + A(A^T)^2$$

Simplify $$A(A^T)^2$$.

$$A(A^T)^2 = (AA^T) \cdot A^T = I \cdot A^T = A^T$$

Final result.

$$\frac{1}{2}A[(A+A^T)^2 + (A-A^T)^2] = A^3 + A^T$$

The correct answer is Option 4: $$A^3 + A^T$$.

We need to find the sum of prime factors of $$|P^{-1}AP - 2I|$$. For any invertible matrix $$P$$, we use the key property that $$|P^{-1}AP - 2I| = |P^{-1}(A - 2I)P| = |P^{-1}||A - 2I||P| = |A - 2I|$$ because $$P^{-1}AP - 2I = P^{-1}AP - P^{-1}(2I)P = P^{-1}(A - 2I)P$$.

The matrix $$A - 2I$$ equals $$\begin{bmatrix} 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}$$. To calculate its determinant, we expand along the first row: $$|A - 2I| = 0 \cdot \begin{vmatrix} 0 & 11 \\ 3 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 11 \\ 3 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 6 & 0 \\ 3 & 3 \end{vmatrix}$$, which simplifies to $$0 - 1(6 \cdot 0 - 11 \cdot 3) + 2(6 \cdot 3 - 0 \cdot 3) = 0 - 1(0 - 33) + 2(18 - 0) = 0 + 33 + 36 = 69$$.

Since $$|P^{-1}AP - 2I| = |A - 2I|$$, it follows that $$|P^{-1}AP - 2I| = 69$$. The prime factorisation of 69 is $$69 = 3 \times 23$$, and thus the sum of its prime factors is $$3 + 23 = 26$$. Therefore, the correct answer is Option 1: 26.

From $$AB^{-1} = A^{-1}$$, multiply by $$A$$ on the right: $$A B^{-1} A = I \implies B = A^2$$.

 Given $$BCB^{-1} = A$$. Since $$B=A^2$$, we have $$A^2 C A^{-2} = A$$.

 This implies $$C$$ is similar to $$A$$ (specifically $$C = A^{-2} A A^2 = A$$). So $$C = A$$.

 Since $$A^2 = B$$, then $$C^2 = B$$.

 Use the Characteristic Equation of $$B$$: $$|B - \lambda I| = 0$$.

$$\begin{vmatrix} 1-\lambda & 3 \\ 1 & 5-\lambda \end{vmatrix} = (1-\lambda)(5-\lambda) - 3 = \lambda^2 - 6\lambda + 2 = 0$$

: By Cayley-Hamilton, $$B^2 - 6B + 2I = O$$. Substitute $$B = C^2$$:

$$(C^2)^2 - 6(C^2) + 2I = 0 \implies C^4 - 6C^2 + 2I = 0$$

\Compare with $$C^4 + \alpha C^2 + \beta I = 0$$: $$\alpha = -6, \beta = 2$$.

 $$2\beta - \alpha = 2(2) - (-6) = 4 + 6 = 10$$.

Correct Option: D (10)

The system of equations is:

$$x + y + z = 4$$ ... (1)

$$2x + 5y + 5z = 17$$ ... (2)

$$x + 2y + mz = n$$ ... (3)

For infinitely many solutions, the system must be consistent and the determinant of the coefficient matrix must be zero.

Find the determinant.

Setting $$D = 0$$: $$3m - 6 = 0 \Rightarrow m = 2$$.

For infinitely many solutions, we also need consistency.

With $$m = 2$$, equation (3) becomes: $$x + 2y + 2z = n$$.

From equations (1) and (2): Subtract 2×(1) from (2): $$3y + 3z = 9 \Rightarrow y + z = 3$$.

From (1): $$x = 4 - (y + z) = 4 - 3 = 1$$. So $$x = 1$$.

Substituting into (3): $$1 + 2y + 2z = n \Rightarrow 1 + 2(y + z) = n \Rightarrow 1 + 6 = n \Rightarrow n = 7$$.

Check which equation is satisfied.

$$m = 2, n = 7$$.

Option (1): $$m^2 + n^2 - mn = 4 + 49 - 14 = 39$$. ✓

The correct answer is Option (1): $$m^2 + n^2 - mn = 39$$.

The given conditions tell us that $$\begin{pmatrix}1\\0\\1\end{pmatrix}$$, $$\begin{pmatrix}-1\\0\\1\end{pmatrix}$$, $$\begin{pmatrix}0\\1\\0\end{pmatrix}$$ are eigenvectors of $$A$$ with eigenvalues 2, 4, 2 respectively.

So $$A - 3I$$ has eigenvalues $$2-3 = -1$$, $$4-3 = 1$$, $$2-3 = -1$$ corresponding to the same eigenvectors.

Since $$\det(A - 3I) = (-1)(1)(-1) = 1 \neq 0$$, the matrix $$A - 3I$$ is invertible.

Therefore the system $$(A - 3I)\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix}$$ has a unique solution.

To verify, let us express $$\begin{pmatrix}1\\2\\3\end{pmatrix}$$ in terms of the eigenvectors. Let:

$$\begin{pmatrix}1\\2\\3\end{pmatrix} = a\begin{pmatrix}1\\0\\1\end{pmatrix} + b\begin{pmatrix}-1\\0\\1\end{pmatrix} + c\begin{pmatrix}0\\1\\0\end{pmatrix}$$

From the second component: $$c = 2$$.

From the first component: $$a - b = 1$$.

From the third component: $$a + b = 3$$.

Solving: $$a = 2, b = 1$$.

Then the solution is:

$$\begin{pmatrix}x\\y\\z\end{pmatrix} = \frac{2}{-1}\begin{pmatrix}1\\0\\1\end{pmatrix} + \frac{1}{1}\begin{pmatrix}-1\\0\\1\end{pmatrix} + \frac{2}{-1}\begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}-2\\0\\-2\end{pmatrix} + \begin{pmatrix}-1\\0\\1\end{pmatrix} + \begin{pmatrix}0\\-2\\0\end{pmatrix} = \begin{pmatrix}-3\\-2\\-1\end{pmatrix}$$

The answer is Option A: unique solution.

A=[[1,2],[0,1]]. adj(A)=[[1,-2],[0,1]]. (adj A)^n=[[1,-2n],[0,1]].

B=I+adj(A)+...+(adj A)^{10}. Sum: [[11, -2(0+1+...+10)],[0,11]]=[[11,-110],[0,11]].

Sum of elements=11-110+0+11=-88.

The answer is Option (3): -88.

We need to verify two statements about the matrix $$f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$.

We observe that this is a rotation matrix representing a rotation by angle $$x$$ about the z-axis.

Let us first analyze the claim that $$f(-x)$$ is the inverse of $$f(x)$$.

First, we compute $$f(-x)$$:

$$ f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Next, we compute the product $$f(x) \cdot f(-x)$$:

$$ f(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Entry (1,1): $$\cos^2 x + \sin^2 x = 1$$

Entry (1,2): $$\cos x \sin x - \sin x \cos x = 0$$

Entry (2,1): $$\sin x \cos x - \cos x \sin x = 0$$

Entry (2,2): $$\sin^2 x + \cos^2 x = 1$$

All other entries work out to give the identity matrix.

$$ f(x) \cdot f(-x) = I_3 $$

Hence, $$f(-x) = [f(x)]^{-1}$$, proving the first statement.

Now consider the second statement: $$f(x)f(y) = f(x+y)$$.

We compute the product $$f(x) \cdot f(y)$$:

$$ f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

Entry (1,1): $$\cos x \cos y - \sin x \sin y = \cos(x+y)$$

Entry (1,2): $$-\cos x \sin y - \sin x \cos y = -\sin(x+y)$$

Entry (2,1): $$\sin x \cos y + \cos x \sin y = \sin(x+y)$$

Entry (2,2): $$-\sin x \sin y + \cos x \cos y = \cos(x+y)$$

$$ f(x) \cdot f(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = f(x+y) $$

Thus, the second statement also holds.

The correct answer is Option (4): Both Statement I and Statement II are true.

We have a 2×2 matrix $$A$$ with $$|A| = 2$$ and the sum of its diagonal elements equal to $$-3$$.

Since for a 2×2 matrix the characteristic equation can be written as $$\lambda^2 - (\text{trace})\,\lambda + \det(A) = 0$$, substituting the given trace $$-3$$ and determinant $$2$$ yields $$\lambda^2 + 3\lambda + 2 = 0$$ which factors as $$(\lambda + 1)(\lambda + 2) = 0$$.

This gives the eigenvalues $$\lambda_1 = -1$$ and $$\lambda_2 = -2$$.

By the Cayley-Hamilton theorem, $$A$$ satisfies its characteristic equation, so $$A^2 + 3A + 2I = O$$.

Comparing this with the general form $$A^2 + xA + yI = O$$ shows that $$x = 3$$ and $$y = 2$$.

Interpreting these values as the semi-major and semi-minor axes gives $$a = x = 3$$ and $$b = y = 2$$, so the equation of the hyperbola becomes $$\frac{X^2}{9} - \frac{Y^2}{4} = 1$$.

Next, since $$c^2 = a^2 + b^2 = 9 + 4 = 13$$, it follows that $$e = \frac{c}{a} = \frac{\sqrt{13}}{3}$$, hence $$e^2 = \frac{13}{9}$$ and $$e^4 = \frac{169}{81}$$.

The length of the latus rectum is given by $$l = \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3}$$, so $$l^2 = \frac{64}{9}$$.

Finally, evaluating $$81(e^4 + l^2)$$ yields $$81\left(\frac{169}{81} + \frac{64}{9}\right) = 81 \times \frac{169}{81} + 81 \times \frac{64}{9},$$ which simplifies to $$= 169 + 9 \times 64 = 169 + 576 = \mathbf{745}$$.

The answer is $$\mathbf{745}$$.

Let $$A = \begin{bmatrix}a&b\\b&d\end{bmatrix}$$ (symmetric). From $$A\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}3\\7\end{bmatrix}$$: $$a+b=3$$ and $$b+d=7$$.

From $$\det(A) = ad - b^2 = 1$$ and $$a = 3-b, d = 7-b$$: $$(3-b)(7-b) - b^2 = 21-10b = 1$$, so $$b=2, a=1, d=5$$.

$$A = \begin{bmatrix}1&2\\2&5\end{bmatrix}$$, $$A^{-1} = \begin{bmatrix}5&-2\\-2&1\end{bmatrix}$$.

From $$A^{-1} = \alpha A + \beta I$$: comparing entries gives $$2\alpha = -2$$ (so $$\alpha = -1$$) and $$\alpha + \beta = 5$$ (so $$\beta = 6$$).

$$\alpha + \beta = -1 + 6 = \boxed{5}$$.

We have $$AB = C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$, so $$B = A^{-1}C$$.

$$|A| = 2(1) - 0 + 1(0 - 1) = 1$$. $$|C| = 1 \times 3 \times 1 = 3$$ (upper triangular).

$$\alpha = |B| = |A^{-1}||C| = 1 \times 3 = 3$$.

Computing $$A^{-1}$$:

$$A^{-1} = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix}$$

$$B = A^{-1}C = \begin{bmatrix} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$$

$$\beta = \text{tr}(B) = 1 + 1 + (-1) = 1$$.

$$\alpha^3 + \beta^3 = 27 + 1 = \boxed{28}$$.

We have $$A = I_2 - 2MM^T$$, where $$M$$ is a $$2 \times 1$$ real matrix with $$M^TM = I_1 = [1]$$, i.e., $$M^TM = 1$$.

Note that $$A$$ is a Householder reflection matrix. We need to find all $$\lambda$$ such that $$AX = \lambda X$$ for some non-zero $$X$$.

Case 1: $$X = M$$

$$AM = (I_2 - 2MM^T)M = M - 2M(M^TM) = M - 2M = -M$$

So $$\lambda = -1$$ is an eigenvalue.

Case 2: $$X$$ perpendicular to $$M$$, i.e., $$M^TX = 0$$.

$$AX = (I_2 - 2MM^T)X = X - 2M(M^TX) = X - 0 = X$$

So $$\lambda = 1$$ is an eigenvalue.

The eigenvalues of $$A$$ are $$\lambda = 1$$ and $$\lambda = -1$$.

Sum of squares = $$1^2 + (-1)^2 = 2$$.

The answer is $$\boxed{2}$$.

A is 2×2 with eigenvalues -1 and 3.

Trace of A = -1 + 3 = 2, det(A) = (-1)(3) = -3.

By Cayley-Hamilton: $$A^2 - 2A - 3I = 0$$, so $$A^2 = 2A + 3I$$.

Trace of $$A^2$$ = 2·trace(A) + 3·trace(I) = 2(2) + 3(2) = 4 + 6 = 10.

The answer is $$\boxed{10}$$.

We are given a $$3 \times 3$$ matrix $$A$$ with non-negative real entries such that:

$$A\begin{bmatrix}1\\1\\1\end{bmatrix} = 3\begin{bmatrix}1\\1\\1\end{bmatrix}$$

This means the sum of each row of $$A$$ equals 3. That is, if $$A = [a_{ij}]$$, then for each row $$i$$: $$a_{i1} + a_{i2} + a_{i3} = 3$$, and all $$a_{ij} \geq 0$$.

We want to maximize $$\det(A)$$.

By the AM-GM inequality applied to matrices, or by direct optimization, the maximum determinant of a doubly stochastic-like matrix with row sums equal to a constant occurs at a specific structure.

The Hadamard inequality states that for a matrix with rows $$\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3$$:

$$|\det(A)| \leq |\mathbf{r}_1||\mathbf{r}_2||\mathbf{r}_3|$$

But this bound is achieved when rows are orthogonal.

Let us try the matrix $$A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$. Then $$\det(A) = 27$$, and each row sums to 3. All entries are non-negative.

Can we do better? The rows must satisfy $$a_{i1} + a_{i2} + a_{i3} = 3$$ with $$a_{ij} \geq 0$$. Each row vector lies in the simplex $$\{(x,y,z) : x+y+z=3, x,y,z \geq 0\}$$, which lies on the plane $$x+y+z=3$$.

The maximum norm of a row vector on this simplex is achieved at the vertices $$(3,0,0)$$, $$(0,3,0)$$, $$(0,0,3)$$, where $$|\mathbf{r}| = 3$$.

For orthogonal rows on this simplex, the best we can do is the identity matrix scaled by 3, giving $$\det = 27$$.

Let us verify this is indeed the maximum. Consider any matrix with row sums 3 and non-negative entries. Write $$A = 3I + B$$ where $$B$$ has row sums 0. Actually, let us use Lagrange multipliers or just check.

Consider the general structure with rows $$\mathbf{r}_i$$ on the simplex. Since the simplex is a subset of the plane $$x+y+z=3$$, the normal to this plane is $$\mathbf{n} = (1,1,1)/\sqrt{3}$$. All row vectors have the same projection onto $$\mathbf{n}$$, namely $$3/\sqrt{3} = \sqrt{3}$$.

Writing each row as $$\mathbf{r}_i = \sqrt{3}\hat{n} + \mathbf{w}_i$$ where $$\mathbf{w}_i \perp \mathbf{n}$$:

$$\det(A) = \det(\sqrt{3}\hat{n}\mathbf{1}^T + W)$$ where $$W$$ has rows $$\mathbf{w}_i$$.

Since $$\hat{n}\mathbf{1}^T$$ has rank 1, and using the matrix determinant lemma approach, the determinant can be expanded. For the $$3I$$ matrix, the rows are $$(3,0,0)$$, $$(0,3,0)$$, $$(0,0,3)$$ which are mutually orthogonal with norm 3 each.

By Hadamard's inequality: $$|\det(A)| \leq 3 \times 3 \times 3 = 27$$.

This bound is achieved by $$A = 3I$$, so the maximum determinant is $$27$$.

The answer is $$27$$.

The given matrices are of the form
$$A=\begin{pmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{pmatrix},$$
where $$a,b,c,d$$ come from the set $$S=\{0,3,5,7,11,13,17,19\}$$ (8 elements).

First find the determinant of $$A$$. Expanding along the third row,

$$\det(A)=0\;C_{31}+5\,C_{32}+0\,C_{33}=5\,(-1)^{3+2}\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$

$$\Rightarrow\quad \det(A)=-5\,(ad-bc).$$

The matrix is invertible $$\iff$$ $$\det(A)\neq0 \iff ad-bc\neq0.$$
Hence we must count the quadruples $$(a,b,c,d)\in S^4$$ for which $$ad\neq bc$$ and subtract this from the total number of quadruples.

Total quadruples: $$|S|^4=8^4=4096.$$

Let $$N$$ be the number of “bad’’ quadruples with $$ad=bc$$. We will find $$N$$ and then compute $$4096-N$$.

For $$ad$$ to be zero, at least one of $$a,d$$ is $$0$$. For $$bc$$ to be zero as well, at least one of $$b,c$$ is $$0$$.

• $$a=0$$ (1 choice) and $$d$$ arbitrary (8 choices):
  Pairs $$(b,c)$$ satisfying $$bc=0$$ are obtained by “at least one zero’’:
  $$(b=0, c\in S)$$ (8) and $$(b\in S, c=0)$$ (8) minus the double-count $$(0,0).$$
  Thus $$15$$ such pairs. So the count here is $$1\times8\times15=120.$$

• $$a\neq0$$ (7 choices) and $$d=0$$ (1 choice): the same 15 pairs for $$(b,c)$$.
  Count $$7\times1\times15=105.$$

Total for Case I: $$120+105=225.$$

Restrict to $$T=S\setminus\{0\}=\{3,5,7,11,13,17,19\}$$ (7 elements).
Put $$p=ad=bc\neq0$$. For every fixed product value $$p$$, let $$n_p$$ be the number of ordered pairs $$(x,y)\in T^2$$ with $$xy=p$$. Then

number of ordered quadruples with product $$p$$ is $$n_p^2$$ (choose $$(a,d)$$ and $$(b,c)$$ independently).

Because all numbers in $$T$$ are distinct primes, the only way two products coincide is by swapping the order of a pair or taking a square:

• Squares: $$(3,3),(5,5),\dots,(19,19)$$ - 7 products, each with $$n_p=1.$$

• Products of two distinct primes: there are $${7\choose2}=21$$ such products; for each, the ordered pairs $$(x,y),(y,x)$$ give $$n_p=2.$$

Hence

$$\sum n_p^2 \;=\;7\cdot1^2 + 21\cdot2^2 = 7 + 84 = 91.$$

Total “bad’’ quadruples: $$N = 225 + 91 = 316.$$

Therefore the number of invertible matrices is

$$4096 - 316 = 3780.$$

Answer: 3780

The entry rule is: $$a_{ij}=1$$ if $$j+1$$ is divisible by $$i$$, otherwise $$a_{ij}=0$$, with $$i,j\in\{1,2,3\}$$.

Step 1: Constructing $$M$$
For each column index $$j$$ the number $$j+1$$ equals 2, 3, 4 respectively. Check divisibility by every row index $$i$$:

$$ \begin{array}{c|ccc} & j=1 & j=2 & j=3\\ j+1 & 2 & 3 & 4\\ \hline i=1 & 1 & 1 & 1\\ i=2 & 1 & 0 & 1\\ i=3 & 0 & 1 & 0 \end{array} $$

Hence
$$ M=\begin{pmatrix} 1&1&1\\ 1&0&1\\ 0&1&0 \end{pmatrix}. $$

Step 2: Determinant of $$M$$
Using the first row expansion,

$$ \det M =1\bigl|\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr| -1\bigl|\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\bigr| +1\bigl|\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\bigr| =1(-1)-1(0)+1(1)=0. $$

Since $$\det M=0$$, the rank of $$M$$ is <3 and $$M$$ is not invertible.

Step 3: Characteristic polynomial and eigenvalues

$$ \begin{aligned} \lvert M-\lambda I\rvert &=\begin{vmatrix} 1-\lambda&1&1\\ 1&-\lambda&1\\ 0&1&-\lambda \end{vmatrix}\\[2mm] &=(1-\lambda)(\lambda^2-1)-1(-\lambda)+1(1)\\ &=-\lambda^3+\lambda^2+2\lambda\\ &=-\lambda\bigl(\lambda^2-\lambda-2\bigr)\\ &=-\lambda(\lambda-2)(\lambda+1). \end{aligned} $$

Thus the eigenvalues are $$\lambda_1=0,\;\lambda_2=2,\;\lambda_3=-1$$.

Step 4: Verifying each option

Option A: $$\det M=0$$, so $$M$$ is not invertible ⇒ Option A is false.

Option B: The eigenvalue $$-1$$ exists, so there is a non-zero vector $$X$$ satisfying $$MX=-X$$. Hence Option B is true.

Option C: Because $$\det M=0$$, the null-space of $$M$$ is non-trivial, i.e. $$\{X\in\mathbb{R}^3:MX=\mathbf{0}\}\neq\{\mathbf{0}\}$$. Option C is true.

Option D: Since $$2$$ is an eigenvalue, $$\det(M-2I)=0$$, so $$M-2I$$ is not invertible. Option D is false.

Final result
Option B and Option C are correct.

Let $$A$$ be symmetric ($$A^T = A$$), and $$B, C$$ be skew-symmetric ($$B^T = -B$$, $$C^T = -C$$).

Properties of powers of symmetric and skew-symmetric matrices.

If $$A$$ is symmetric, then $$A^n$$ is symmetric for all $$n$$: $$(A^n)^T = (A^T)^n = A^n$$.

If $$B$$ is skew-symmetric, then $$B^2$$ is symmetric: $$(B^2)^T = (B^T)^2 = (-B)^2 = B^2$$.

So $$B^{26} = (B^2)^{13}$$ is symmetric.

If $$C$$ is skew-symmetric, then $$C^{13} = C \cdot (C^2)^6$$. Since $$C^2$$ is symmetric, $$(C^{13})^T = ((C^2)^6)^T \cdot C^T = (C^2)^6 \cdot (-C) = -C^{13}$$. So $$C^{13}$$ is skew-symmetric.

Analyze S1: $$A^{13}B^{26} - B^{26}A^{13}$$.

Let $$P = A^{13}$$ (symmetric) and $$Q = B^{26}$$ (symmetric).

$$(PQ - QP)^T = (PQ)^T - (QP)^T = Q^T P^T - P^T Q^T = QP - PQ = -(PQ - QP)$$

So $$A^{13}B^{26} - B^{26}A^{13}$$ is skew-symmetric, not symmetric.

S1 is false.

Analyze S2: $$A^{26}C^{13} - C^{13}A^{26}$$.

Let $$P = A^{26}$$ (symmetric) and $$R = C^{13}$$ (skew-symmetric).

$$(PR - RP)^T = R^T P^T - P^T R^T = (-R)P - P(-R) = -RP + PR = PR - RP$$

So $$A^{26}C^{13} - C^{13}A^{26}$$ is symmetric.

S2 is true.

Conclusion.

Only S2 is true.

The correct answer is Option A: Only S2 is true.

We are given that $$A$$ and $$B$$ are non-zero $$n \times n$$ matrices satisfying $$A^2 + B = A^2B$$.

Rewriting the given equation, we have $$A^2 + B = A^2B$$, which implies $$A^2 = A^2B - B = (A^2 - I)B$$. It follows that $$A^2(I - B) = -B$$.

If $$(I - B)$$ is invertible, then from $$A^2(I - B) = -B$$ we deduce $$A^2 = -B(I - B)^{-1} = B(B - I)^{-1}$$. By Cayley-Hamilton, $$(B - I)^{-1}$$ is a polynomial in $$B$$ and hence commutes with $$B$$. Therefore,

$$A^2B = B(B - I)^{-1}B = B \cdot B(B - I)^{-1} = B \cdot A^2 = BA^2$$.

The correct answer is Option D: $$A^2B = BA^2$$.

Given $$A = \frac{1}{2}\begin{bmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}$$, we can recognize this matrix as a rotation matrix by noting that $$A = \begin{bmatrix} \cos 60° & \sin 60° \\ -\sin 60° & \cos 60° \end{bmatrix}$$, which is precisely $$R(-60°)$$, a rotation by $$-60°$$.

Since $$A = R(-60°)$$, it follows in general that $$A^n = R(-60n°)$$ for any integer $$n$$.

Applying this to the thirtieth power gives $$A^{30} = R(-60° \times 30) = R(-1800°) = R(0°) = I$$, because $$-1800° = -5 \times 360°$$.

Similarly, for the twenty-fifth power one finds $$A^{25} = R(-60° \times 25) = R(-1500°) = R(-1500° + 1440°) = R(-60°) = A$$.

Checking the proposed relations, Option A asserts $$A^{30} - A^{25} = I - A = 2I$$, which would force $$A = -I$$ and is false. Option B asserts $$A^{30} + A^{25} + A = I + A + A = I + 2A \neq I$$, which also fails. Option C states $$A^{30} + A^{25} - A = I + A - A = I$$, which holds true. Option D claims $$A^{30} = A^{25}$$ or $$I = A$$, which is false.

Therefore the correct answer is Option C.

We are given: $$A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$$, where $$i = \sqrt{-1}$$. Also $$M = A^T B A$$ and we need to find the inverse of $$AM^{2023}A^T$$.

First we verify that $$A$$ is an orthogonal matrix by computing its transpose and multiplying: $$A^T = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$$ and then $$A^T A = \begin{bmatrix} \frac{1}{10} + \frac{9}{10} & \frac{3}{10} - \frac{3}{10} \\ \frac{3}{10} - \frac{3}{10} & \frac{9}{10} + \frac{1}{10} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$. Hence $$A^T = A^{-1}$$.

Next, since $$M = A^T B A$$, we have $$M^2 = (A^T B A)(A^T B A) = A^T B (A A^T) B A = A^T B^2 A$$. By induction it follows that $$M^n = A^T B^n A$$ for any positive integer $$n$$, and in particular $$M^{2023} = A^T B^{2023} A$$.

Multiplying by $$A$$ on the left and by $$A^T$$ on the right gives $$AM^{2023}A^T = A (A^T B^{2023} A) A^T = (A A^T) B^{2023} (A A^T) = I \, B^{2023} \, I = B^{2023}$$.

To find $$B^{2023}$$, note that $$B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$$ is upper triangular with 1s on the diagonal. The power of such a matrix is $$B^n = \begin{bmatrix} 1 & -ni \\ 0 & 1 \end{bmatrix}$$, which can be checked by computing $$B^2 = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2i \\ 0 & 1 \end{bmatrix}$$ and proceeding by induction. Thus $$B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}$$.

Finally, the inverse of a $$2\times2$$ upper triangular matrix $$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$$ is $$\begin{bmatrix} 1 & -a \\ 0 & 1 \end{bmatrix}$$ because $$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & -a \\ 0 & 1 \end{bmatrix} = I$$. Here $$a = -2023i$$, so the inverse of $$B^{2023}$$ is $$\begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}$$. Therefore, $$(AM^{2023}A^T)^{-1} = \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}$$.

Given: $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$$. Find the sum of diagonal elements of $$(A + I)^{11}$$.

Adding the identity matrix to $$A$$ gives $$A + I = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & -1 \\ 0 & 12 & -2 \end{pmatrix}.$$

The matrix has a block structure: the (1,1) entry is 2 and the lower-right 2×2 block is $$B = \begin{pmatrix} 5 & -1 \\ 12 & -2 \end{pmatrix}.$$ Hence $$(A+I)^{11} = \begin{pmatrix} 2^{11} & 0 & 0 \\ 0 & & \\ 0 & B^{11} & \end{pmatrix}$$ and its trace is $$2^{11} + \mathrm{tr}(B^{11}).$$

The characteristic polynomial of $$B$$ is $$\det(B - \lambda I) = (5-\lambda)(-2-\lambda) + 12 = \lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2),$$ so the eigenvalues are $$\lambda_1 = 1, \; \lambda_2 = 2.$$

Therefore $$\mathrm{tr}(B^{11}) = \lambda_1^{11} + \lambda_2^{11} = 1 + 2^{11} = 1 + 2048 = 2049.$$

It follows that $$\mathrm{tr}((A+I)^{11}) = 2^{11} + 2049 = 2048 + 2049 = 4097.$$

The correct answer is Option (3): $$\boxed{4097}$$.

Given: $$A$$ is a $$2 \times 2$$ matrix with all non-zero entries and $$A^2 = I$$.

Let $$a = \text{tr}(A)$$ and $$b = |A| = \det(A)$$.

To begin, Since $$A^2 = I$$, we have $$\det(A^2) = \det(I) = 1$$, so $$(\det A)^2 = 1$$, giving $$b = \pm 1$$.

Next, Taking trace of $$A^2 = I$$: $$\text{tr}(A^2) = \text{tr}(I) = 2$$.

For a $$2 \times 2$$ matrix: $$\text{tr}(A^2) = (\text{tr}A)^2 - 2\det(A)$$.

$$ a^2 - 2b = 2 $$

From here, If $$b = 1$$: $$a^2 = 4$$, so $$a = \pm 2$$. This gives eigenvalues both equal to 1 (or both -1), meaning $$A = \pm I$$. But then off-diagonal entries would be 0, contradicting $$a_{ij} \neq 0$$.

If $$b = -1$$: $$a^2 = 0$$, so $$a = 0$$. The eigenvalues are $$+1$$ and $$-1$$, and the matrix can have all non-zero entries. ✓

Continuing,

$$ 3a^2 + 4b^2 = 3(0) + 4(1) = 4 $$

The correct answer is 4.

Given $$A' = \alpha A + I$$ where A is a 2×2 real matrix.

Let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, then $$A' = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$$.

From $$A' = \alpha A + I$$:

$$a = \alpha a + 1 \Rightarrow a = \frac{1}{1-\alpha}$$

$$c = \alpha b$$ and $$b = \alpha c$$

$$d = \alpha d + 1 \Rightarrow d = \frac{1}{1-\alpha}$$

From $$b = \alpha c$$ and $$c = \alpha b$$: $$b = \alpha^2 b$$, so $$b(1-\alpha^2) = 0$$. Since $$\alpha \neq \pm 1$$, we get $$b = 0$$ and $$c = 0$$.

Therefore $$A = kI$$ where $$k = \frac{1}{1-\alpha}$$.

$$A^2 - A = k^2I - kI = (k^2 - k)I$$

$$\det(A^2 - A) = (k^2 - k)^2 = 4$$

$$k^2 - k = \pm 2$$

Case 1: $$k^2 - k - 2 = 0 \Rightarrow (k-2)(k+1) = 0$$

$$k = 2 \Rightarrow \alpha = \frac{1}{2}$$ or $$k = -1 \Rightarrow \alpha = 2$$

Case 2: $$k^2 - k + 2 = 0$$

Discriminant $$= 1 - 8 = -7 < 0$$. No real solutions.

Sum of all possible values of $$\alpha = \frac{1}{2} + 2 = \frac{5}{2}$$

The equation $$(a-c)x^2 + (b-a)x + (c-b) = 0$$ has $$x = 1$$ as a root (since the sum of coefficients is zero), and by Vieta’s formulas the other root is $$\alpha = \frac{c-b}{a-c}$$.

The given matrix is singular for either root; in particular, when $$\alpha = 1$$ the first two rows coincide, so the determinant vanishes.

Setting $$p = a - c$$, $$q = b - a$$ and $$r = c - b$$, one observes that $$p + q + r = 0$$. The expression then becomes $$\frac{p^2}{qr} + \frac{q^2}{pr} + \frac{r^2}{pq} = \frac{p^3 + q^3 + r^3}{pqr}$$. Since $$p + q + r = 0$$, the identity $$p^3 + q^3 + r^3 = 3pqr$$ applies, giving $$\frac{3pqr}{pqr} = 3$$.

The correct answer is Option 2: 3.

We are given that $$P^2 = I - P$$, $$P^\alpha + P^\beta = \gamma I - 29P$$, and $$P^\alpha - P^\beta = \delta I - 13P$$, where $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$.

To express the powers of $$P$$ in the form $$aI + bP$$, observe that $$P^2 = I - P$$. Assuming $$P^{n-1} = a_{n-1}I + b_{n-1}P$$, one finds $$P^n = P\cdot P^{n-1} = a_{n-1}P + b_{n-1}P^2 = a_{n-1}P + b_{n-1}(I - P) = b_{n-1}I + (a_{n-1}-b_{n-1})P.$$

Computing these coefficients successively gives $$P^1 = 0\cdot I + 1\cdot P,\quad P^2 = 1\cdot I + (-1)\cdot P,\quad P^3 = (-1)I + 2P,\quad P^4 = 2I + (-3)P,\quad P^5 = (-3)I + 5P,\quad P^6 = 5I + (-8)P,\quad P^7 = (-8)I + 13P,\quad P^8 = 13I + (-21)P.$$

From $$P^\alpha + P^\beta = \gamma I - 29P$$ the coefficient of $$P$$ yields $$b_\alpha + b_\beta = -29$$, and from $$P^\alpha - P^\beta = \delta I - 13P$$ the coefficient of $$P$$ gives $$b_\alpha - b_\beta = -13$$. Solving these equations leads to $$b_\alpha = -21$$ and $$b_\beta = -8$$, which correspond to $$\alpha = 8$$ and $$\beta = 6$$.

The coefficients of $$I$$ in the same expressions are then $$\gamma = a_8 + a_6 = 13 + 5 = 18$$ and $$\delta = a_8 - a_6 = 13 - 5 = 8$$.

Hence, $$\alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = \boxed{24}$$. The correct answer is Option D.

We need to find the sum of all elements of the matrix $$\sum_{n=1}^{50} B^n$$.

Let $$P = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}$$ and $$Q = \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$$. Then $$B = PAQ$$.

We verify that $$PQ = QP = I$$ (so $$Q = P^{-1}$$):

$$PQ = \begin{bmatrix} 1(-1)+2(1) & 1(-2)+2(1) \\ -1(-1)+(-1)(1) & -1(-2)+(-1)(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

Therefore $$B^n = PA^nQ$$.

$$A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$$ is an upper triangular matrix with ones on the diagonal, so:

$$A^n = \begin{bmatrix} 1 & \frac{n}{51} \\ 0 & 1 \end{bmatrix}$$

$$\sum_{n=1}^{50} A^n = \begin{bmatrix} 50 & \frac{1}{51}\sum_{n=1}^{50} n \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & \frac{1275}{51} \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & 25 \\ 0 & 50 \end{bmatrix}$$

First: $$P \cdot \begin{bmatrix} 50 & 25 \\ 0 & 50 \end{bmatrix} = \begin{bmatrix} 50 & 125 \\ -50 & -75 \end{bmatrix}$$

Then: $$\begin{bmatrix} 50 & 125 \\ -50 & -75 \end{bmatrix} \cdot \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 75 & 25 \\ -25 & 25 \end{bmatrix}$$

$$75 + 25 + (-25) + 25 = 100$$

The correct answer is Option D: $$100$$.

We need to find the number of $$5 \times 5$$ matrices with entries from $$\{0, 1\}$$ such that each row sum and each column sum equals 1.

A matrix with entries 0 and 1 where every row and every column sums to 1 must have exactly one 1 in each row and exactly one 1 in each column. Such a matrix is called a permutation matrix.

Each permutation matrix of order $$n$$ corresponds to a unique permutation of $$\{1, 2, 3, 4, 5\}$$:
the 1 in row 1 can be in any of 5 columns,
the 1 in row 2 can be in any of the remaining 4 columns,
the 1 in row 3 can be in any of the remaining 3 columns,
and so on. This shows that the number of such matrices is $$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$.

The answer is Option B: 120.

To solve this quickly, use the properties of determinants and adjoints for an $$n \times n$$ matrix (here, $$m = 3$$):

Expand the determinant of $$A$$:

$$|A| = 2(4 - 1) - 1(2 - 0) + 0 = 2(3) - 2 = 4$$

For a $$3 \times 3$$ matrix, $$|kA| = k^3|A|$$:

$$|2A| = 2^3 \cdot 4 = 8 \cdot 4 = 32 = 2^5$$

The formula for $$|\text{adj}(\text{adj}(\dots(\text{adj}(M))\dots))|$$ with $$k$$ adjoints is $$|M|^{(m-1)^k}$$.

Here, $$k = 3$$ (three adjoints) and $$m = 3$$:

$$| \text{adj}(\text{adj}(\text{adj}(2A))) | = |2A|^{(3-1)^3} = |2A|^{2^3} = |2A|^8$$

Substitute the value of $$|2A|$$:

$$(2^5)^8 = 2^{40}$$

The problem states this equals $$(16)^n$$:

$$(2^4)^n = 2^{4n}$$

Equating powers:

$$4n = 40 \implies n = 10$$

Observe that $$P$$ is an orthogonal matrix ($$P P^T = I$$), which means $$P^T = P^{-1}$$.

The expression $$P^T Q^{2007} P$$ simplifies using the property $$Q = PAP^T$$:

$$P^T (P A P^T)^{2007} P = P^T (P A^{2007} P^T) P = (P^T P) A^{2007} (P^T P) = A^{2007}$$

For $$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$, the $$n$$-th power is $$A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$.

Thus, $$A^{2007} = \begin{bmatrix} 1 & 2007 \\ 0 & 1 \end{bmatrix}$$, giving:

• $$a = 1$$
• $$b = 2007$$
• $$c = 0$$
• $$d = 1$$

Final Calculation:

$$2a + b - 3c - 4d = 2(1) + 2007 - 3(0) - 4(1) = 2 + 2007 - 4 = \mathbf{2005}$$

Correct Option: (B)

We are given $$A = \begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix}$$, $$A^{-1} = \alpha A + \beta I$$, and $$\alpha + \beta = -2$$. We need to find $$4\alpha^2 + \beta^2 + \lambda^2$$.

Compute $$A^{-1}$$.

$$\det(A) = 1 \times 10 - 5 \times \lambda = 10 - 5\lambda$$

$$A^{-1} = \frac{1}{10 - 5\lambda}\begin{bmatrix} 10 & -5 \\ -\lambda & 1 \end{bmatrix}$$

Compute $$\alpha A + \beta I$$.

$$\alpha A + \beta I = \alpha\begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix} + \beta\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \alpha + \beta & 5\alpha \\ \lambda\alpha & 10\alpha + \beta \end{bmatrix}$$

Equate and solve.

Comparing the (1,1) entries:

$$\alpha + \beta = \frac{10}{10 - 5\lambda}$$

Since $$\alpha + \beta = -2$$:

$$-2 = \frac{10}{10 - 5\lambda} \implies -2(10 - 5\lambda) = 10 \implies -20 + 10\lambda = 10 \implies \lambda = 3$$

So $$\det(A) = 10 - 15 = -5$$.

Comparing the (1,2) entries:

$$5\alpha = \frac{-5}{-5} = 1 \implies \alpha = \frac{1}{5}$$

From $$\alpha + \beta = -2$$:

$$\beta = -2 - \frac{1}{5} = -\frac{11}{5}$$

Verify with other entries.

Entry (2,1): $$\lambda\alpha = 3 \times \frac{1}{5} = \frac{3}{5}$$ and $$\frac{-\lambda}{\det(A)} = \frac{-3}{-5} = \frac{3}{5}$$. ✔

Entry (2,2): $$10\alpha + \beta = \frac{10}{5} - \frac{11}{5} = -\frac{1}{5}$$ and $$\frac{1}{\det(A)} = \frac{1}{-5} = -\frac{1}{5}$$. ✔

Compute the required expression.

$$4\alpha^2 + \beta^2 + \lambda^2 = 4 \times \frac{1}{25} + \frac{121}{25} + 9 = \frac{4 + 121}{25} + 9 = \frac{125}{25} + 9 = 5 + 9 = 14$$

The answer is Option C: 14.

Let $$A = \begin{bmatrix} p & q \\ q & r \end{bmatrix}$$ (since $$A$$ is symmetric) with $$|A| = pr - q^2 = 2$$.

We have $$\begin{bmatrix} 2 & 1 \\ 3 & \frac{3}{2} \end{bmatrix} \begin{bmatrix} p & q \\ q & r \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$$.

Multiplying the matrices yields $$\begin{bmatrix} 2p + q & 2q + r \\ 3p + \frac{3q}{2} & 3q + \frac{3r}{2} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ \alpha & \beta \end{bmatrix}$$.

From the first row we have $$2p + q = 1$$ and $$2q + r = 2$$, so $$q = 1 - 2p$$ and $$r = 2 - 2q = 2 - 2(1 - 2p) = 4p$$.

Using the condition $$|A| = pr - q^2 = 2$$ gives $$p(4p) - (1 - 2p)^2 = 2$$, which simplifies to $$4p^2 - 1 + 4p - 4p^2 = 2$$ and hence $$4p - 1 = 2$$, so $$p = \frac{3}{4}$$. Therefore $$q = -\frac{1}{2}$$ and $$r = 3$$.

From the second row we find $$\alpha = 3 \cdot \frac{3}{4} + \frac{3}{2} \cdot \left(-\frac{1}{2}\right) = \frac{9}{4} - \frac{3}{4} = \frac{3}{2}$$ and $$\beta = 3 \cdot \left(-\frac{1}{2}\right) + \frac{3}{2} \cdot 3 = -\frac{3}{2} + \frac{9}{2} = 3$$.

The sum of diagonal elements is $$s = \frac{3}{4} + 3 = \frac{15}{4}$$, and thus $$\frac{\beta s}{\alpha^2} = \frac{3 \cdot \frac{15}{4}}{\left(\frac{3}{2}\right)^2} = \frac{\frac{45}{4}}{\frac{9}{4}} = \frac{45}{9} = \boxed{5}$$.

Let $$A = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}$$ with $$A^3 = A$$. We need to find the positive value of $$a$$ and determine $$n$$ such that $$a \in (n-1, n]$$.

The condition $$A^3 = A$$ implies $$A^3 - A = 0$$, so $$A(A^2 - I) = 0$$. Hence the minimal polynomial of $$A$$ divides $$x^3 - x = x(x-1)(x+1)$$ and the eigenvalues of $$A$$ can only be $$0,1,-1$$.

Computing $$A^2$$ gives $$A^2 = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix},$$ and the row computations yield Row 1: $$(0+a+2, 0+0+2c, 0+3+0) = (a+2,2c,3),$$ Row 2: $$(0+0+3, a+0+3c, 2a+0+0) = (3,a+3c,2a),$$ Row 3: $$(0+ac+0, 1+0+0, 2+3c+0) = (ac,1,2+3c).$$ Therefore $$A^2 = \begin{pmatrix} a+2 & 2c & 3 \\ 3 & a+3c & 2a \\ ac & 1 & 2+3c \end{pmatrix}.$$

Next $$A^3 = A^2\cdot A,$$ and multiplying out gives the entries Row 1: $$(a+2)(0)+2c(a)+3(1),\;(a+2)(1)+2c(0)+3(c),\;(a+2)(2)+2c(3)+3(0) = (2ac+3,\;a+2+3c,\;2a+4+6c),$$ Row 2: $$3(0)+(a+3c)(a)+2a(1),\;3(1)+(a+3c)(0)+2a(c),\;3(2)+(a+3c)(3)+2a(0) = (a^2+3ac+2a,\;3+2ac,\;6+3a+9c),$$ Row 3: $$ac(0)+1(a)+(2+3c)(1),\;ac(1)+1(0)+(2+3c)(c),\;ac(2)+1(3)+(2+3c)(0) = (a+2+3c,\;ac+2c+3c^2,\;2ac+3).$$

Setting $$A^3 = A$$ and comparing with $$A = \begin{pmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{pmatrix}$$ gives from the (1,1) entry $$2ac+3=0\implies ac=-\dfrac{3}{2},$$ from (1,2) $$a+2+3c=1\implies a+3c=-1,$$ from (1,3) $$2a+4+6c=2\implies a+3c=-1$$ (same), and from (2,1) $$a^2+3ac+2a=a\implies a^2+a-\dfrac{9}{2}=0.$$

Rewriting $$a^2+a-\dfrac{9}{2}=0$$ as $$2a^2+2a-9=0$$ leads to $$a=\dfrac{-2\pm\sqrt{4+72}}{4}=\dfrac{-2\pm\sqrt{76}}{4}=\dfrac{-2\pm2\sqrt{19}}{4}=\dfrac{-1\pm\sqrt{19}}{2},$$ so the positive value is $$a=\dfrac{-1+\sqrt{19}}{2}.$$ Since $$\sqrt{19}\approx4.359$$, we have $$a\approx1.68.$$

Because $$a\approx1.68\in(1,2]$$, it follows that $$n-1=1$$ and hence $$n=2$$. Therefore the answer is $$2$$.

Write the given matrix in the form $$M = I + N$$, where $$I$$ is the identity matrix and $$N$$ is easy to handle.

First find $$I$$ minus $$M$$:
$$M = \begin{pmatrix}\dfrac52 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac12\end{pmatrix}, \quad I = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}.$$

Hence
$$N = M - I = \begin{pmatrix}\dfrac52-1 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac12-1\end{pmatrix} = \begin{pmatrix}\dfrac32 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac32\end{pmatrix} = \dfrac32\begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix}.$$

Check that $$N$$ is nilpotent of index 2:
Let $$A = \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix}$$. Then
$$A^2 = \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix} \begin{pmatrix}1 & 1 \\ -1 & -1\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.$$
Therefore $$N^2 = \left(\dfrac32\right)^2 A^2 = 0.$$

For any positive integer $$k$$, $$(I+N)^k = I + kN$$ because the binomial expansion stops at the linear term when $$N^2 = 0.$$

Thus
$$M^{2022} = (I+N)^{2022} = I + 2022N.$$

Compute $$2022N$$:
$$2022N = 2022 \times \begin{pmatrix}\dfrac32 & \dfrac32 \\[4pt] -\dfrac32 & -\dfrac32\end{pmatrix} = \begin{pmatrix}2022 \times \dfrac32 & 2022 \times \dfrac32 \\[4pt] -2022 \times \dfrac32 & -2022 \times \dfrac32\end{pmatrix} = \begin{pmatrix}3033 & 3033 \\[4pt] -3033 & -3033\end{pmatrix}.$$

Add $$I$$ to finish:
$$M^{2022} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} + \begin{pmatrix}3033 & 3033 \\ -3033 & -3033\end{pmatrix} = \begin{pmatrix}3034 & 3033 \\ -3033 & -3032\end{pmatrix}.$$

Hence $$M^{2022} = \begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}.$$

Option A which is: $$\begin{pmatrix} 3034 & 3033 \\ -3033 & -3032 \end{pmatrix}$$

The given matrix is
$$A=\begin{pmatrix}\beta & 0 & 1\\[2pt] 2 & 1 & -2\\[2pt] 3 & 1 & -2\end{pmatrix}\qquad(\beta\in\mathbb{R}).$$

The matrix in the question is
$$M=A^{7}-(\beta-1)A^{6}-\beta A^{5}=A^{5}\Bigl[A^{2}-(\beta-1)A-\beta I\Bigr].$$
Hence
$$\det M=(\det A)^{5}\,\det\!\Bigl[A^{2}-(\beta-1)A-\beta I\Bigr].$$
$$M$$ is singular if and only if at least one of the two determinants on the right‐hand side is zero.

1. Determinant of $$A$$
We first compute $$\det(A-\lambda I)$$ to obtain both $$\det A$$ and the characteristic polynomial.

$$ A-\lambda I=\begin{pmatrix} \beta-\lambda & 0 & 1\\ 2 & 1-\lambda & -2\\ 3 & 1 & -2-\lambda \end{pmatrix}. $$

Expanding along the first row:
$$ \det(A-\lambda I)=(\beta-\lambda) \det\begin{pmatrix}1-\lambda & -2\\ 1 & -2-\lambda\end{pmatrix} +1\!\det\begin{pmatrix}2 & 1-\lambda\\ 3 & 1\end{pmatrix}. $$

$$ \det\begin{pmatrix}1-\lambda & -2\\ 1 & -2-\lambda\end{pmatrix} =(1-\lambda)(-2-\lambda)+2 =-\bigl(1-\lambda\bigr)(2+\lambda)+2 =\lambda+\lambda^{2}. $$

$$ \det\begin{pmatrix}2 & 1-\lambda\\ 3 & 1\end{pmatrix} =2\cdot1-3(1-\lambda)= -1+3\lambda. $$

Therefore
$$ \det(A-\lambda I)=\lambda(1+\lambda)(\beta-\lambda)+3\lambda-1 =-\lambda^{3}+(\beta-1)\lambda^{2}+(\beta+3)\lambda-1. $$

The constant term gives $$\det A=-1\neq0,$$ which is independent of $$\beta$$. Hence $$\det A^{5}=(-1)^{5}=-1\neq0$$ for every real $$\beta$$. The only possible reason for $$M$$ to be singular is therefore

2. $$\det\!\bigl[A^{2}-(\beta-1)A-\beta I\bigr]=0$$.

Write the characteristic polynomial in monic form (changing sign):
$$ f(\lambda)=\lambda^{3}-(\beta-1)\lambda^{2}-(\beta+3)\lambda+1=0. \qquad -(1) $$

If $$\lambda$$ is an eigenvalue of $$A$$, then $$\lambda$$ satisfies $$(1)$$. Define the quadratic polynomial coming from the second factor in $$M$$:
$$ g(\lambda)=\lambda^{2}-(\beta-1)\lambda-\beta. \qquad -(2) $$

The determinant in question vanishes exactly when some eigenvalue $$\lambda$$ of $$A$$ satisfies both $$(1)$$ and $$(2)$$. Thus we need a common root of the two polynomials. Apply the Euclidean algorithm:

Divide $$f(\lambda)$$ by $$g(\lambda)$$:

Since $$\lambda\cdot g(\lambda)=\lambda^{3}-(\beta-1)\lambda^{2}-\beta\lambda,$$
$$ f(\lambda)-\lambda g(\lambda)= \bigl[-(\beta+3)\lambda+1\bigr]-\bigl[-\beta\lambda\bigr] =-3\lambda+1. $$

The remainder is $$-3\lambda+1$$. A common root must therefore satisfy
$$ -3\lambda+1=0\;\Longrightarrow\;\lambda=\tfrac13. \qquad -(3) $$

Substituting $$\lambda=\tfrac13$$ into $$(2)$$ gives
$$ \left(\tfrac13\right)^{2}-(\beta-1)\left(\tfrac13\right)-\beta=0, $$ $$ \frac19-\frac{\beta-1}{3}-\beta=0. $$

Multiply by 9:
$$ 1-3(\beta-1)-9\beta=0\;\Longrightarrow\;4-12\beta=0, $$ $$ \beta=\frac13. $$

Hence $$M$$ is singular only for $$\beta=\dfrac13$$, and the asked quantity is
$$ 9\beta=9\cdot\frac13=3. $$

Final Answer: 3

We are given that A is a $$3 \times 3$$ symmetric matrix (so $$A^T = A$$) and B is a $$3 \times 3$$ skew-symmetric matrix (so $$B^T = -B$$). We need to find which statement is NOT true.

Option 1: $$A^4 - B^4$$ is symmetric. We check: $$(A^4 - B^4)^T = (A^T)^4 - (B^T)^4 = A^4 - (-B)^4 = A^4 - B^4$$. So $$A^4 - B^4$$ is symmetric. This is TRUE.

Option 2: $$AB - BA$$ is symmetric. We check: $$(AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T = (-B)(A) - (A)(-B) = -BA + AB = AB - BA$$. So $$AB - BA$$ is symmetric. This is TRUE.

Option 3: $$B^5 - A^5$$ is skew-symmetric. We check: $$(B^5 - A^5)^T = (B^T)^5 - (A^T)^5 = (-B)^5 - A^5 = -B^5 - A^5 = -(B^5 + A^5)$$. For this to be skew-symmetric, we need $$(B^5 - A^5)^T = -(B^5 - A^5) = -B^5 + A^5$$. But we got $$-B^5 - A^5$$. These are equal only if $$A^5 = -A^5$$, i.e., $$A^5 = 0$$, which is not generally true. So this is NOT TRUE.

Option 4: $$AB + BA$$ is skew-symmetric. We check: $$(AB + BA)^T = B^T A^T + A^T B^T = (-B)(A) + (A)(-B) = -BA - AB = -(AB + BA)$$. So $$AB + BA$$ is skew-symmetric. This is TRUE.

Hence, the correct answer is Option 3.

We have a $$2 \times 2$$ matrix A with $$\det(A) = -1$$ and $$\det((A + I)(\text{Adj}(A) + I)) = 4$$.

For a $$2 \times 2$$ matrix, $$\text{Adj}(A) = \det(A) \cdot A^{-1} = -A^{-1}$$.

$$(A + I)(\text{Adj}(A) + I) = (A + I)(-A^{-1} + I) = (A + I)(I - A^{-1})$$

$$= A \cdot I - A \cdot A^{-1} + I \cdot I - I \cdot A^{-1}$$

$$= A - I + I - A^{-1} = A - A^{-1}$$

Let $$\text{tr}(A) = s$$. By Cayley-Hamilton: $$A^2 - sA + \det(A) \cdot I = 0$$, so $$A^2 = sA + I$$.

From $$A \cdot A^{-1} = I$$: $$A^{-1} = \frac{1}{\det(A)}(sI - A) = -(sI - A) = A - sI$$.

$$A - A^{-1} = A - (A - sI) = sI$$

$$\det(A - A^{-1}) = \det(sI) = s^2 = 4$$

$$s = \pm 2$$

The sum of diagonal elements of A can be $$2$$ or $$-2$$. From the options, $$2$$ is available.

Therefore, the correct answer is Option B: 2.

We are given $$A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$$, and we wish to determine the nature of $$MN^2$$ where $$M = \sum_{k=1}^{10} A^{2k}$$ and $$N = \sum_{k=1}^{10} A^{2k-1}$$.

Since $$A^2 = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} -4 & 0 \\ 0 & -4 \end{pmatrix} = -4I$$, it follows that $$A^{2k} = (-4)^k I$$ for any integer $$k\ge1$$.

Substituting this into the definition of $$M$$ gives $$M = \sum_{k=1}^{10} (-4)^k I = \left[\sum_{k=1}^{10}(-4)^k\right]I$$. Let $$m = \sum_{k=1}^{10}(-4)^k = \frac{-4((-4)^{10} - 1)}{-4 - 1} = \frac{-4(4^{10} - 1)}{-5} = \frac{4(4^{10} - 1)}{5}$$, so that $$M = mI$$ where $$m = \frac{4(4^{10}-1)}{5}$$.

Similarly, since $$A^{2k-1} = A^{2(k-1)}A = (-4)^{k-1}A$$, one finds $$N = \sum_{k=1}^{10}(-4)^{k-1}A = \left[\sum_{k=0}^{9}(-4)^k\right]A$$. Defining $$n = \sum_{k=0}^{9}(-4)^k = \frac{1-(-4)^{10}}{1-(-4)} = \frac{1-4^{10}}{5} = -\frac{4^{10}-1}{5}$$ yields $$N = nA$$.

Proceeding to $$N^2$$, we have $$N^2 = n^2A^2 = n^2(-4I) = -4n^2I$$. Consequently, $$MN^2 = (mI)(-4n^2I) = -4mn^2I$$, which is a scalar multiple of the identity matrix.

Using $$m = \frac{4(4^{10}-1)}{5}$$ and $$n^2 = \frac{(4^{10}-1)^2}{25}$$, it follows that $$-4mn^2 = -4\cdot\frac{4(4^{10}-1)}{5}\cdot\frac{(4^{10}-1)^2}{25} = \frac{-16(4^{10}-1)^3}{125}$$, which is clearly not equal to 1. Therefore, $$MN^2$$ is a non-identity scalar matrix. Since any scalar matrix $$cI$$ satisfies $$(cI)^T = cI$$, it is symmetric.

Hence, the correct classification is that $$MN^2$$ is a non-identity symmetric matrix, corresponding to Option A.

We start with the matrix $$A = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$$ and check which option cannot be obtained by a single elementary row operation.

Option A: $$\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix}$$. Apply $$R_1 \to R_1 + R_2$$: $$(-1+1,\; 2+(-1)) = (0, 1)$$, and $$R_2$$ stays $$(1, -1)$$. This gives $$\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix}$$. Obtainable.

Option B: $$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$$. Apply $$R_1 \leftrightarrow R_2$$ (row swap): rows get interchanged, giving $$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$$. Obtainable.

Option C: $$\begin{pmatrix} -1 & 2 \\ -2 & 7 \end{pmatrix}$$. Row 1 is unchanged, so the operation must be on $$R_2$$. We need $$R_2 \to R_2 + kR_1$$: $$(1 + k(-1),\; -1 + 2k) = (-2, 7)$$. From the first entry: $$1 - k = -2$$, so $$k = 3$$. From the second entry: $$-1 + 6 = 5 \neq 7$$. So no value of $$k$$ works. We could also try $$R_2 \to cR_2$$: $$(c, -c) = (-2, 7)$$ gives $$c = -2$$ and $$c = -7$$, contradiction. No single elementary row operation produces this matrix. NOT obtainable.

Option D: $$\begin{pmatrix} -1 & 2 \\ -1 & 3 \end{pmatrix}$$. Row 1 unchanged, so $$R_2 \to R_2 + kR_1$$: $$(1-k,\; -1+2k) = (-1, 3)$$. From the first: $$k = 2$$. From the second: $$-1+4 = 3$$. Both consistent. Obtainable with $$R_2 \to R_2 + 2R_1$$.

Hence, the correct answer is Option C: $$\begin{pmatrix} -1 & 2 \\ -2 & 7 \end{pmatrix}$$.

We have two $$3 \times 3$$ non-zero real matrices $$A$$ and $$B$$ such that $$AB = O$$ (the zero matrix). We need to determine which statement is correct.

Since $$B$$ is a non-zero matrix, there exists at least one column of $$B$$, say $$\mathbf{b}_j$$, that is a non-zero vector. Now, $$AB = O$$ means that $$A\mathbf{b}_j = \mathbf{0}$$ for every column $$\mathbf{b}_j$$ of $$B$$. In particular, $$A\mathbf{b}_j = \mathbf{0}$$ with $$\mathbf{b}_j \neq \mathbf{0}$$.

This shows that the homogeneous system $$AX = \mathbf{0}$$ has a non-trivial solution (namely $$X = \mathbf{b}_j$$). A homogeneous system that admits a non-trivial solution must have infinitely many solutions (since any scalar multiple of a non-trivial solution is also a solution). Hence Option B is correct.

We can also verify that the other options are incorrect. Since $$AX = \mathbf{0}$$ has a non-trivial solution, $$\det(A) = 0$$, which means $$A$$ is singular. Consequently, $$\text{adj}(A)$$ is not invertible (because for a singular $$3 \times 3$$ matrix, $$\det(\text{adj}(A)) = (\det A)^2 = 0$$). This rules out Options A and D. Also, since $$AB = O$$ and $$A \neq O$$, if $$B$$ were invertible we could multiply on the right by $$B^{-1}$$ to get $$A = O$$, a contradiction. So $$B$$ is not invertible, ruling out Option C.

Hence, the correct answer is Option B.

We need to find the value of $$A'BA$$ where $$A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$B = \begin{pmatrix} 9^2 & -10^2 & 11^2 \\ 12^2 & 13^2 & -14^2 \\ -15^2 & 16^2 & 17^2 \end{pmatrix}$$.

$$A$$ is $$3 \times 1$$, so $$A'$$ (transpose) is $$1 \times 3$$.

$$B$$ is $$3 \times 3$$.

$$A'BA$$ is $$(1 \times 3)(3 \times 3)(3 \times 1) = 1 \times 1$$ (a scalar).

$$BA = \begin{pmatrix} 81 & -100 & 121 \\ 144 & 169 & -196 \\ -225 & 256 & 289 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$

Row 1: $$81 - 100 + 121 = 102$$

Row 2: $$144 + 169 - 196 = 117$$

Row 3: $$-225 + 256 + 289 = 320$$

$$BA = \begin{pmatrix} 102 \\ 117 \\ 320 \end{pmatrix}$$

$$A'BA = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 102 \\ 117 \\ 320 \end{pmatrix} = 102 + 117 + 320 = 539$$

Therefore, the correct answer is Option D: $$539$$.

Consider the $$2 \times 2$$ matrix $$A = \begin{pmatrix} 4 & -2 \\ \alpha & \beta \end{pmatrix}$$. Its trace is $$\text{tr}(A) = 4 + \beta$$ and its determinant is $$\det(A) = 4\beta - (-2)\alpha = 4\beta + 2\alpha$$. The characteristic equation is given by $$\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0 \implies \lambda^2 - (4+\beta)\lambda + (4\beta + 2\alpha) = 0$$.

By the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation, we have $$A^2 - (4+\beta)A + (4\beta + 2\alpha)I = O$$. Since we are also given $$A^2 + \gamma A + 18I = O$$, comparing coefficients of $$A$$ and $$I$$ yields $$\gamma = -(4+\beta)$$ and $$18 = 4\beta + 2\alpha = \det(A)$$.

Therefore, $$\det(A) = \boxed{18}$$, and the answer is Option B.

We can write $$A = I + N$$ where $$N = \begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$$.

Since $$N$$ is strictly upper triangular, it is nilpotent: $$N^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and $$N^3 = O$$.

Because $$I$$ and $$N$$ commute, by the binomial theorem: $$A^n = (I + N)^n = I + nN + \binom{n}{2}N^2$$ for all $$n \geq 0$$.

The $$(1,3)$$ entry of $$A^n$$ is $$0 + n \cdot 0 + \binom{n}{2} \cdot 1 = \dfrac{n(n-1)}{2}$$.

Now $$B = 7A^{20} - 20A^7 + 2I$$, so $$b_{13} = 7 \cdot \dfrac{20 \cdot 19}{2} - 20 \cdot \dfrac{7 \cdot 6}{2} + 2 \cdot 0 = 7 \cdot 190 - 20 \cdot 21 = 1330 - 420 = 910$$.

The correct answer is Option A: $$910$$.

We are given $$A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ and $$B_0 = A^{49} + 2A^{98}$$, with $$B_n = \text{Adj}(B_{n-1})$$ for $$n \geq 1$$.

Step 1: Find powers of A.

Matrix A swaps the first two rows (it is a permutation matrix). Therefore:

$$A^2 = I$$ (the identity matrix)

For any integer $$n$$: $$A^{\text{odd}} = A$$ and $$A^{\text{even}} = I$$.

Since 49 is odd: $$A^{49} = A$$

Since 98 is even: $$A^{98} = I$$

Step 2: Compute $$B_0$$.

$$B_0 = A + 2I = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$

Step 3: Find $$\det(B_0)$$.

Expanding along the third row:

$$\det(B_0) = 3 \cdot (4 - 1) = 3 \times 3 = 9 = 3^2$$

Step 4: Use the adjugate determinant formula.

For an $$n \times n$$ matrix M: $$\det(\text{Adj}(M)) = [\det(M)]^{n-1}$$

For our $$3 \times 3$$ matrices ($$n = 3$$): $$\det(B_n) = [\det(B_{n-1})]^2$$

Step 5: Compute $$\det(B_4)$$ iteratively.

$$\det(B_0) = 3^2$$

$$\det(B_1) = (3^2)^2 = 3^4$$

$$\det(B_2) = (3^4)^2 = 3^8$$

$$\det(B_3) = (3^8)^2 = 3^{16}$$

$$\det(B_4) = (3^{16})^2 = 3^{32}$$

The correct answer is Option C: $$3^{32}$$

We are given $$A = \begin{pmatrix} 2 & -1 \\ 0 & 2 \end{pmatrix}$$ and $$B = I - {}^5C_1(\text{adj } A) + {}^5C_2(\text{adj } A)^2 - \ldots - {}^5C_5(\text{adj } A)^5$$. For a 2x2 matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, the adjugate is $$\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$, so $$\text{adj } A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$$.

Recognizing the binomial expansion, we write $$B = \sum_{k=0}^{5} {}^5C_k (-\text{adj } A)^k = (I - \text{adj } A)^5$$. Then $$I - \text{adj } A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$$.

Let $$M = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix} = -\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$. Since $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$$, we have $$M^5 = (-1)^5 \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}$$, hence $$B = \begin{pmatrix} -1 & -5 \\ 0 & -1 \end{pmatrix}$$. Summing its entries gives $$-1 + (-5) + 0 + (-1) = -7$$.

Therefore, the answer is Option C: $$\textbf{-7}$$.

Given $$A = \begin{pmatrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$ and $$A^n = \begin{pmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{pmatrix}$$.

Write $$A = I + N$$ where $$N = \begin{pmatrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}$$. Since $$N^2 = \begin{pmatrix} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ and $$N^3 = 0$$, we have

Matching entries gives $$na = 48$$ so $$a = \frac{48}{n}$$, and $$nb = 96$$ so $$b = \frac{96}{n}$$. Also, from the $$(1,3)$$ entry, $$na + \binom{n}{2}ab = 2160$$.

Substituting yields

which simplifies to

Thus $$a = \frac{48}{12} = 4$$ and $$b = \frac{96}{12} = 8$$, and hence $$n + a + b = 12 + 4 + 8 = 24$$.

The answer is $$\boxed{24}$$.

We have $$A = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$$ and $$B = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}$$.

$$A^2 = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 4-2 & -4+2 \\ 2-1 & -2+1 \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} = A$$. Since $$A^2 = A$$, we have $$A^n = A$$ for all $$n \geq 1$$ (that is, $$A$$ is idempotent).

$$B^2 = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1-2 & -2+4 \\ 1-2 & -2+4 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = B$$. Since $$B^2 = B$$, we have $$B^m = B$$ for all $$m \geq 1$$.

Since $$A^n = A$$ and $$B^m = B$$, the equation $$nA^n + mB^m = I$$ becomes $$nA + mB = I$$, namely

$$n\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} + m\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ which gives the system:
$$2n - m = 1 \quad \cdots (1)$$
$$-2n + 2m = 0 \quad \cdots (2)$$
$$n - m = 0 \quad \cdots (3)$$
$$-n + 2m = 1 \quad \cdots (4)$$

From equation (3) we have $$n = m$$. Substituting into (1) gives $$2n - n = 1 \implies n = 1$$, and checking in (4) yields $$-1 + 2(1) = 1$$ ✓.

Therefore, the only solution is $$(n, m) = (1, 1)$$, which lies in $$\{1,2,\ldots,10\}^2$$. Hence, the number of elements in the set is $$\boxed{1}$$.

We are given the matrices $$M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}$$ and $$N = \sum_{k=1}^{49} M^{2k}$$, together with the relation $$(I - M^2)N = -2I\,. $$

We first compute $$M^2$$ by multiplying the given matrix by itself: $$M^2 = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} = \begin{bmatrix} -\alpha^2 & 0 \\ 0 & -\alpha^2 \end{bmatrix} = -\alpha^2 I\,. $$ From this it follows that $$M^{2k} = (M^2)^k = (-\alpha^2)^k I = (-1)^k \alpha^{2k} I\,. $$

Consequently, the sum defining $$N$$ becomes $$N = \sum_{k=1}^{49}(-1)^k \alpha^{2k} I = \Bigl(\sum_{k=1}^{49}(-\alpha^2)^k\Bigr) I\,. $$ This is a geometric series with first term $$-\alpha^2$$ and common ratio $$-\alpha^2$$, so we use the formula $$\sum_{k=1}^{49}(-\alpha^2)^k = \frac{-\alpha^2\bigl(1-(-\alpha^2)^{49}\bigr)}{1-(-\alpha^2)} = \frac{-\alpha^2(1+\alpha^{98})}{1+\alpha^2}\,. $$

Next, since $$I - M^2 = I + \alpha^2 I = (1+\alpha^2)I$$, substituting into $$(I - M^2)N = -2I$$ gives $$(1+\alpha^2)\cdot \frac{-\alpha^2(1+\alpha^{98})}{1+\alpha^2} = -2\,, $$ which simplifies to $$-\alpha^2(1+\alpha^{98}) = -2\,. $$ Hence we obtain $$\alpha^2(1+\alpha^{98}) = 2 \quad\Longrightarrow\quad \alpha^2 + \alpha^{100} = 2\,. $$

Finally, testing positive integers shows that for $$\alpha = 1$$ we have $$1 + 1 = 2$$ ✔ whereas for $$\alpha = 2$$ we get $$4 + 2^{100} \gg 2$$ ✘. Therefore, the only positive integral solution is $$\alpha = \mathbf{1}\,. $$

We have $$X = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$ and $$A = \begin{pmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{pmatrix}$$, and we seek $$k \in \mathbb{N}$$ such that $$X'A^kX = 33$$.

Rather than computing $$A^k$$ in closed form, we observe that $$X'A^kX$$ can be computed iteratively since $$A^kX = A(A^{k-1}X)$$. Let us define $$V_k = A^kX$$ and track the sequence $$X'V_k$$.

We compute $$V_1 = AX = \begin{pmatrix} -1+2+3 \\ 0+1+6 \\ 0+0-1 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \\ -1 \end{pmatrix}$$, giving $$X'V_1 = 4 + 7 - 1 = 10$$.

Now $$V_2 = AV_1 = \begin{pmatrix} -4+14-3 \\ 0+7-6 \\ 0+0+1 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ 1 \end{pmatrix}$$, giving $$X'V_2 = 9$$.

Continuing, $$V_3 = AV_2 = \begin{pmatrix} -7+2+3 \\ 0+1+6 \\ 0+0-1 \end{pmatrix} = \begin{pmatrix} -2 \\ 7 \\ -1 \end{pmatrix}$$, giving $$X'V_3 = 4$$. Next, $$V_4 = AV_3 = \begin{pmatrix} 2+14-3 \\ 0+7-6 \\ 0+0+1 \end{pmatrix} = \begin{pmatrix} 13 \\ 1 \\ 1 \end{pmatrix}$$, giving $$X'V_4 = 15$$.

A clear pattern emerges. For even $$k$$, $$V_k$$ has the form $$\begin{pmatrix} c \\ 1 \\ 1 \end{pmatrix}$$, and for odd $$k$$, $$V_k = \begin{pmatrix} d \\ 7 \\ -1 \end{pmatrix}$$. At each even step, the first component increases by 6: $$c_2 = 7, c_4 = 13, c_6 = 19, \ldots$$, so $$c_{2m} = 6m + 1$$. This gives $$X'A^{2m}X = (6m + 1) + 1 + 1 = 6m + 3$$.

Setting $$6m + 3 = 33$$ yields $$m = 5$$, so $$k = 2m = 10$$. We can verify the odd-$$k$$ formula gives $$X'A^{2m+1}X = -6m + 10$$, which never equals 33 for any natural number $$m$$, confirming that $$k = 10$$ is the unique solution.

Hence, the correct answer is $$\boxed{10}$$.

We have $$A = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}$$, $$B = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}$$.

Part 1: Finding $$\alpha_1$$ from $$(A+B)^2 = A^2 + \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$$

Expand $$(A+B)^2 = A^2 + AB + BA + B^2$$.

So the condition becomes $$AB + BA + B^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$$.

$$AB = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}\begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \beta - 1 & 1 \\ 2\beta + \alpha & 2 \end{pmatrix}$$

$$BA = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix} = \begin{pmatrix} \beta + 2 & -\beta + \alpha \\ 1 & -1 \end{pmatrix}$$

$$B^2 = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \beta^2 + 1 & \beta \\ \beta & 1 \end{pmatrix}$$

$$AB + BA + B^2 = \begin{pmatrix} (\beta-1) + (\beta+2) + (\beta^2+1) & 1 + (-\beta+\alpha) + \beta \\ (2\beta+\alpha) + 1 + \beta & 2 + (-1) + 1 \end{pmatrix}$$

$$= \begin{pmatrix} \beta^2 + 2\beta + 2 & \alpha + 1 \\ 3\beta + \alpha + 1 & 2 \end{pmatrix}$$

From position $$(1,2)$$: $$\alpha + 1 = 2 \implies \alpha_1 = 1$$.

From position $$(1,1)$$: $$\beta^2 + 2\beta + 2 = 2 \implies \beta^2 + 2\beta = 0 \implies \beta(\beta + 2) = 0$$.

So $$\beta = 0$$ or $$\beta = -2$$.

From position $$(2,1)$$: $$3\beta + \alpha_1 + 1 = 2 \implies 3\beta + 2 = 2 \implies \beta = 0$$.

Therefore $$\alpha_1 = 1$$ (with $$\beta = 0$$).

Part 2: Finding $$\alpha_2$$ from $$(A+B)^2 = B^2$$

$$(A+B)^2 = B^2$$ means $$A^2 + AB + BA = O$$ (the zero matrix).

$$A^2 = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix} = \begin{pmatrix} 1-2 & -1-\alpha \\ 2+2\alpha & -2+\alpha^2 \end{pmatrix} = \begin{pmatrix} -1 & -1-\alpha \\ 2+2\alpha & \alpha^2-2 \end{pmatrix}$$

$$= \begin{pmatrix} -1 + (\beta-1) + (\beta+2) & (-1-\alpha) + 1 + (-\beta+\alpha) \\ (2+2\alpha) + (2\beta+\alpha) + 1 & (\alpha^2-2) + 2 + (-1) \end{pmatrix}$$

$$= \begin{pmatrix} 2\beta & -\beta \\ 3\alpha + 2\beta + 3 & \alpha^2 - 1 \end{pmatrix}$$

From $$(1,1)$$: $$2\beta = 0 \implies \beta = 0$$.

From $$(1,2)$$: $$-\beta = 0 \implies \beta = 0$$. (Consistent.)

From $$(2,2)$$: $$\alpha^2 - 1 = 0 \implies \alpha = 1$$ or $$\alpha = -1$$.

From $$(2,1)$$: $$3\alpha + 0 + 3 = 0 \implies \alpha = -1$$.

Therefore $$\alpha_2 = -1$$.

$$|\alpha_1 - \alpha_2| = |1 - (-1)| = 2$$.

The answer is 2.

We have $$A = \begin{pmatrix} 1+i & 1 \\ -i & 0 \end{pmatrix}$$ and need to find the number of $$n \in \{1, 2, \ldots, 100\}$$ such that $$A^n = A$$.

To find the eigenvalues of $$A$$, we write the characteristic equation:

$$\det(A - \lambda I) = (1+i-\lambda)(0-\lambda) - (1)(-i) = 0$$

$$-\lambda(1+i-\lambda) + i = 0$$

$$\lambda^2 - (1+i)\lambda + i = 0$$

Using the quadratic formula:

$$\lambda = \frac{(1+i) \pm \sqrt{(1+i)^2 - 4i}}{2} = \frac{(1+i) \pm \sqrt{2i - 4i}}{2} = \frac{(1+i) \pm \sqrt{-2i}}{2}$$

Now $$-2i = 2e^{-i\pi/2}$$, so $$\sqrt{-2i} = \sqrt{2}\,e^{-i\pi/4} = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$$.

$$\lambda_1 = \frac{(1+i) + (1-i)}{2} = 1, \qquad \lambda_2 = \frac{(1+i) - (1-i)}{2} = i$$

Next, since the eigenvalues $$\lambda_1 = 1$$ and $$\lambda_2 = i$$ are distinct, $$A$$ is diagonalizable:

$$A = PDP^{-1}, \quad D = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$$

Therefore, $$A^n = PD^nP^{-1}$$, where $$D^n = \begin{pmatrix} 1 & 0 \\ 0 & i^n \end{pmatrix}$$.

In order for $$A^n = A$$, we require $$D^n = D$$, which means:

$$1^n = 1 \quad \text{(always true)}, \quad \text{and} \quad i^n = i$$

$$i^n = i$$ holds when $$n \equiv 1 \pmod{4}$$.

Finally, we count the valid values of $$n$$.

We need $$n \in \{1, 2, \ldots, 100\}$$ with $$n \equiv 1 \pmod{4}$$.

These are: $$n = 1, 5, 9, 13, \ldots, 97$$.

This is an arithmetic sequence with first term 1, common difference 4, and last term 97.

Number of terms: $$\frac{97 - 1}{4} + 1 = 25$$.

The answer is $$\boxed{25}$$.

Given $$A = \begin{pmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}$$, $$B = A - I$$, and $$\omega = \dfrac{\sqrt{3}i - 1}{2}$$. Find the number of $$n \in \{1, 2, \ldots, 100\}$$ such that $$A^n + (\omega B)^n = A + B$$.

Subtracting the identity matrix from $$A$$ gives:

Since every row of $$B$$ is $$(1, -1, -1)$$, the matrix $$B$$ has rank 1.

Multiplying $$B$$ by itself yields:

Thus $$B^2 = -B$$, which implies $$B^3 = B$$. More generally, $$B^n = B$$ if $$n$$ is odd and $$B^n = -B$$ if $$n$$ is even (for $$n \ge 1$$).

Using the relation $$A = I + B$$ together with $$B^2 = -B$$, we compute:

Therefore $$A^2 = A$$, and by induction $$A^n = A$$ for all $$n \ge 1$$.

Considering $$(\omega B)^n$$, we note that:

When $$n$$ is odd, $$B^n = B$$ so $$(\omega B)^n = \omega^n B$$.

When $$n$$ is even, $$B^n = -B$$ so $$(\omega B)^n = -\omega^n B$$.

Substituting into the equation $$A^n + (\omega B)^n = A + B$$ and using $$A^n = A$$ leads to:

In the case of odd $$n$$, this requires $$\omega^n B = B \implies \omega^n = 1$$.

In the case of even $$n$$, it requires $$-\omega^n B = B \implies \omega^n = -1$$.

Since $$\omega = \dfrac{-1 + \sqrt{3}i}{2} = e^{i\cdot 2\pi/3}$$ is a primitive cube root of unity, we have $$\omega^3 = 1$$.

For odd $$n$$, the condition $$\omega^n = 1$$ implies $$n \equiv 0 \pmod{3}$$, and combined with oddness gives $$n \equiv 3 \pmod{6}$$.

Within the set $$\{1, \ldots, 100\}$$, these values are $$n = 3, 9, 15, \ldots, 99$$, whose number is $$\dfrac{99 - 3}{6} + 1 = 17$$.

For even $$n$$, there is no solution because $$\omega^n$$ can only be $$1, \omega, \omega^2$$, none of which equals $$-1$$.

Hence there are 17 such values of $$n$$.

First, note that $$X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$, $$Y = \alpha I + \beta X + \gamma X^2$$, and $$Y^{-1} = \begin{bmatrix} 1/5 & -2/5 & 1/5 \\ 0 & 1/5 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix}$$, and we seek $$(\alpha - \beta + \gamma)^2$$.

Next, compute $$X^2$$: $$X^2 = X \cdot X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Since $$X^3 = 0$$ (the zero matrix), higher powers vanish.

Now, write $$Y$$ explicitly as $$Y = \alpha\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \beta\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} + \gamma\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix}.$$

Since for an upper triangular matrix with constant diagonal $$\alpha$$ the inverse has diagonal entries $$1/\alpha$$, and given that $$Y^{-1}$$ has diagonal entries $$1/5$$, it follows that $$\alpha = 5$$.

Next, the inverse of $$Y$$ can be expressed as $$Y^{-1} = \frac{1}{\alpha}I - \frac{\beta}{\alpha^2}X + \frac{\beta^2 - \alpha\gamma}{\alpha^3}X^2,$$ so its $$(1,2)$$ entry is $$-\frac{\beta}{\alpha^2}$$. Equating this to the given value $$-2/5$$ leads to $$-\frac{\beta}{\alpha^2} = -\frac{2}{5} \implies \beta = \frac{2\alpha^2}{5} = 10.$$

Substituting $$\alpha = 5$$ and $$\beta = 10$$ into the formula for the $$(1,3)$$ entry, $$\frac{\beta^2 - \alpha\gamma}{\alpha^3}$$, and setting it equal to the given $$1/5$$ yields $$\frac{100 - 5\gamma}{125} = \frac{1}{5} \implies 100 - 5\gamma = 25 \implies \gamma = 15.$$

Therefore, $$\alpha - \beta + \gamma = 5 - 10 + 15 = 10$$ and $$(\alpha - \beta + \gamma)^2 = 10^2 = 100,$$ which is the required answer.

We have the set $$S = \left\{\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} : a, b \in \{1, 2, \ldots, 100\}\right\}$$.

Define $$T_n = \{A \in S : A^{n(n+1)} = I\}$$. We need to find $$\left|\bigcap_{n=1}^{100} T_n\right|$$.

For $$A = \begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix}$$, we first compute $$A^2$$:

$$A^2 = \begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix}\begin{pmatrix} -1 & a \\ 0 & b \end{pmatrix} = \begin{pmatrix} 1 & a(-1+b) \\ 0 & b^2 \end{pmatrix}$$

For $$A^m = I$$, the eigenvalues of $$A$$ must satisfy $$\lambda^m = 1$$. The eigenvalues of $$A$$ are $$-1$$ and $$b$$.

Condition on $$(-1)$$: $$(-1)^m = 1$$ requires $$m$$ to be even. Since $$n(n+1)$$ is always even (product of consecutive integers), this condition is automatically satisfied for all $$n$$.

Condition on $$b$$: $$b^{n(n+1)} = 1$$ for all $$n = 1, 2, \ldots, 100$$. Since $$b$$ is a positive integer, the only solution is $$b = 1$$.

To confirm: if $$b \geq 2$$, then $$b^{n(n+1)} \geq 2^2 = 4 \neq 1$$. So $$b = 1$$ is required.

Verifying $$A^2 = I$$ when $$b = 1$$:

$$A^2 = \begin{pmatrix} 1 & a(-1+1) \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$$

Since $$A^2 = I$$, for any even exponent $$m$$: $$A^m = (A^2)^{m/2} = I^{m/2} = I$$.

Since $$n(n+1)$$ is always even, $$A^{n(n+1)} = I$$ holds for all $$n$$.

Therefore, the condition is satisfied for $$b = 1$$ and any $$a \in \{1, 2, \ldots, 100\}$$.

The number of elements in $$\bigcap_{n=1}^{100} T_n$$ is $$100$$.

The correct answer is $$100$$.

We have $$A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$$ and $$AA^T = I_2$$.

Computing $$AA^T = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}\begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^2 & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^2 + \beta^2 \end{bmatrix}$$.

Setting this equal to $$I_2$$, from the $$(1,1)$$ entry: $$1 + \alpha^2 = 1$$, so $$\alpha^2 = 0$$, giving $$\alpha = 0$$.

From the $$(2,2)$$ entry: $$\alpha^2 + \beta^2 = 1$$, so $$\beta^2 = 1$$, giving $$\beta = \pm 1$$.

Therefore, $$\alpha^4 + \beta^4 = 0 + 1 = 1$$.

We are given a $$3 \times 3$$ real matrix $$A = [a_{ij}]$$ where each row sums to 1: $$a_{i1} + a_{i2} + a_{i3} = 1$$ for $$i = 1, 2, 3$$.

Let $$\mathbf{e} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$. The row-sum condition means $$A\mathbf{e} = \mathbf{e}$$.

Then $$A^2 \mathbf{e} = A(A\mathbf{e}) = A\mathbf{e} = \mathbf{e}$$, and similarly $$A^3 \mathbf{e} = \mathbf{e}$$.

This means each row of $$A^3$$ also sums to 1. The sum of ALL entries of $$A^3$$ is the sum of the three row sums = $$1 + 1 + 1 = 3$$.

The answer is $$3$$, which is Option C.

We are given that $$A$$ is a $$3 \times 3$$ symmetric matrix (so $$A^T = A$$) and $$B$$ is a $$3 \times 3$$ skew-symmetric matrix (so $$B^T = -B$$).

Let $$M = A^2B^2 - B^2A^2$$. We compute the transpose of $$M$$: $$M^T = (A^2B^2)^T - (B^2A^2)^T = (B^2)^T(A^2)^T - (A^2)^T(B^2)^T$$.

Since $$A^T = A$$, we have $$(A^2)^T = (A^T)^2 = A^2$$. Since $$B^T = -B$$, we have $$(B^2)^T = (B^T)^2 = (-B)^2 = B^2$$.

Therefore, $$M^T = B^2 \cdot A^2 - A^2 \cdot B^2 = -(A^2B^2 - B^2A^2) = -M$$. This shows that $$M$$ is a skew-symmetric matrix.

For any $$n \times n$$ skew-symmetric matrix with $$n$$ odd, we have $$\det(M) = \det(M^T) = \det(-M) = (-1)^n \det(M)$$. Since $$n = 3$$ is odd, $$\det(M) = -\det(M)$$, which gives $$\det(M) = 0$$.

Since $$\det(M) = 0$$, the matrix $$M$$ is singular. The homogeneous system $$MX = O$$ always has the trivial solution $$X = O$$, and because the determinant is zero, there also exist non-trivial solutions. Hence the system has infinitely many solutions.

Therefore, the system $$(A^2B^2 - B^2A^2)X = O$$ has infinitely many solutions.

We have the matrix

$$A=\begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix}.$$

To understand high powers of $$A$$, we first observe what happens when we multiply $$A$$ by itself. Let us denote $$A^n=\begin{bmatrix}1&0&0\\a_n&b_n&b_n\\1&0&0\end{bmatrix}$$ for some numbers $$a_n$$ and $$b_n$$ (we shall soon find a pattern for them). We start with

$$A^1=A=\begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix},$$

so here $$a_1=0$$ and $$b_1=1$$.

Now we multiply $$A^n$$ by $$A$$ to get $$A^{n+1}$$:

$$A^{n+1}=A^nA =\begin{bmatrix}1&0&0\\a_n&b_n&b_n\\1&0&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix}.$$

We compute each entry row-by-column.

For the first row we get

$$[1,0,0]\times A=[1,0,0],$$

so the first row of every power remains $$[1,0,0]$$. The third row is identical to the first, hence it also stays $$[1,0,0]$$ for every power.

For the second row we multiply $$[a_n,b_n,b_n]$$ with $$A$$:

First column: $$a_n\cdot1+b_n\cdot0+b_n\cdot1=a_n+b_n,$$

Second column: $$a_n\cdot0+b_n\cdot1+b_n\cdot0=b_n,$$

Third column: $$a_n\cdot0+b_n\cdot1+b_n\cdot0=b_n.$$

Hence

$$a_{n+1}=a_n+b_n,\qquad b_{n+1}=b_n.$$

From the initial value $$b_1=1$$ and the recurrence $$b_{n+1}=b_n$$, we see immediately that

$$b_n=1\quad\text{for all }n\ge1.$$

Substituting $$b_n=1$$ into $$a_{n+1}=a_n+1$$ with $$a_1=0$$ gives

$$a_n=n-1.$$ Therefore, for every integer $$n\ge1$$,

$$A^n=\begin{bmatrix} 1&0&0\\ n-1&1&1\\ 1&0&0 \end{bmatrix}.$$

Now we can evaluate the two required powers.

For $$n=2025$$ we get

$$A^{2025}=\begin{bmatrix} 1&0&0\\ 2025-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 2024&1&1\\ 1&0&0 \end{bmatrix}.$$

For $$n=2020$$ we get

$$A^{2020}=\begin{bmatrix} 1&0&0\\ 2020-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 2019&1&1\\ 1&0&0 \end{bmatrix}.$$

Their difference is

$$A^{2025}-A^{2020}= \begin{bmatrix} 1-1&0-0&0-0\\ 2024-2019&1-1&1-1\\ 1-1&0-0&0-0 \end{bmatrix} =\begin{bmatrix} 0&0&0\\ 5&0&0\\ 0&0&0 \end{bmatrix}.$$

Next, we compare this result with the matrices that appear in the answer options. Using the same general formula with $$n=6$$ and $$n=1$$, we have

$$A^6=\begin{bmatrix} 1&0&0\\ 6-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 5&1&1\\ 1&0&0 \end{bmatrix},$$

and obviously

$$A=\begin{bmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{bmatrix}.$$

Subtracting, we find

$$A^6-A= \begin{bmatrix} 1-1&0-0&0-0\\ 5-0&1-1&1-1\\ 1-1&0-0&0-0 \end{bmatrix} =\begin{bmatrix} 0&0&0\\ 5&0&0\\ 0&0&0 \end{bmatrix}.$$

This is exactly the same matrix as $$A^{2025}-A^{2020}$$. Therefore,

$$A^{2025}-A^{2020}=A^6-A.$$

Hence, the correct answer is Option A.

Any matrix $$A$$ can be written as $$P + Q$$ where $$P = \frac{A + A^T}{2}$$ (symmetric) and $$Q = \frac{A - A^T}{2}$$ (skew-symmetric).

With $$A = \begin{bmatrix}2 & 3 \\ a & 0\end{bmatrix}$$, we get: $$P = \begin{bmatrix}2 & \frac{3+a}{2} \\ \frac{3+a}{2} & 0\end{bmatrix}, \quad Q = \begin{bmatrix}0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0\end{bmatrix}.$$

Computing $$\det(Q)$$: $$\det(Q) = 0 \cdot 0 - \frac{3-a}{2} \cdot \frac{a-3}{2} = \frac{(3-a)^2}{4} = 9 \implies (3-a)^2 = 36 \implies 3-a = \pm 6.$$

So $$a = -3$$ or $$a = 9$$.

Computing $$\det(P) = 2 \cdot 0 - \left(\frac{3+a}{2}\right)^2 = -\frac{(3+a)^2}{4}$$.

For $$a = -3$$: $$\det(P) = -\frac{0}{4} = 0$$.

For $$a = 9$$: $$\det(P) = -\frac{144}{4} = -36$$.

The sum of all possible values of $$\det(P)$$ is $$0 + (-36) = -36$$. The modulus of this sum is $$\boxed{36}$$.

We have $$A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$$. First, we compute $$A^2 = A \cdot A = \begin{bmatrix} i^2 + i^2 & -i^2 - i^2 \\ -i^2 - i^2 & i^2 + i^2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}$$.

Let $$B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$, so $$A^2 = -2B$$. We note that $$B^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 2B$$. Therefore $$A^4 = (A^2)^2 = 4B^2 = 8B$$ and $$A^8 = (A^4)^2 = 64B^2 = 128B = \begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}$$.

The system $$A^8 \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ becomes $$128(x - y) = 8$$ and $$128(-x + y) = 64$$, which gives $$x - y = \frac{1}{16}$$ and $$-x + y = \frac{1}{2}$$.

Adding these two equations: $$0 = \frac{1}{16} + \frac{1}{2} = \frac{9}{16}$$, which is a contradiction. Therefore, the system has no solution.

We check whether the relation $$ARB \iff PAP^{-1} = B$$ for some non-singular $$P$$ is an equivalence relation.

Reflexive: Taking $$P = I$$ (the identity matrix), we get $$IAI^{-1} = A$$, so $$ARA$$ holds for every matrix $$A$$.

Symmetric: If $$ARB$$, then $$PAP^{-1} = B$$ for some non-singular $$P$$. This gives $$A = P^{-1}BP$$, which means $$QBQ^{-1} = A$$ where $$Q = P^{-1}$$ is also non-singular. So $$BRA$$ holds.

Transitive: If $$ARB$$ and $$BRC$$, then $$PAP^{-1} = B$$ and $$QBQ^{-1} = C$$ for non-singular $$P$$ and $$Q$$. Substituting, $$C = Q(PAP^{-1})Q^{-1} = (QP)A(QP)^{-1}$$. Since $$QP$$ is non-singular, $$ARC$$ holds.

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation.

We have the matrix

$$A=\begin{bmatrix}0&2\\K&-1\end{bmatrix}$$

and the condition

$$A\bigl(A^{3}+3I\bigr)=2I.$$

First we must compute successive powers of $$A$$. The square of a 2 × 2 matrix is obtained by ordinary matrix multiplication. So

$$A^{2}=A\cdot A =\begin{bmatrix}0&2\\K&-1\end{bmatrix} \begin{bmatrix}0&2\\K&-1\end{bmatrix}.$$

Carrying out the multiplication element-wise:

$$ \begin{aligned} A^{2}_{11}&=0\cdot0+2\cdot K=2K,\\ A^{2}_{12}&=0\cdot2+2\cdot(-1)=-2,\\ A^{2}_{21}&=K\cdot0+(-1)\cdot K=-K,\\ A^{2}_{22}&=K\cdot2+(-1)\cdot(-1)=2K+1. \end{aligned} $$

Hence

$$A^{2}=\begin{bmatrix}2K&-2\\-K&2K+1\end{bmatrix}.$$

Next, $$A^{3}=A^{2}\cdot A$$, so

$$A^{3}= \begin{bmatrix}2K&-2\\-K&2K+1\end{bmatrix} \begin{bmatrix}0&2\\K&-1\end{bmatrix}.$$

Again multiplying entry by entry,

$$ \begin{aligned} A^{3}_{11}&=2K\cdot0+(-2)\cdot K=-2K,\\[4pt] A^{3}_{12}&=2K\cdot2+(-2)\cdot(-1)=4K+2,\\[4pt] A^{3}_{21}&=-K\cdot0+(2K+1)\cdot K=K(2K+1),\\[4pt] A^{3}_{22}&=-K\cdot2+(2K+1)\cdot(-1)=-2K-(2K+1)=-4K-1. \end{aligned} $$

Thus

$$A^{3}=\begin{bmatrix}-2K&4K+2\\K(2K+1)&-4K-1\end{bmatrix}.$$

Now we form $$A^{3}+3I$$. Because the identity matrix is $$I=\begin{bmatrix}1&0\\0&1\end{bmatrix},$$ we have

$$3I=\begin{bmatrix}3&0\\0&3\end{bmatrix},$$

so that

$$A^{3}+3I= \begin{bmatrix}-2K+3&4K+2\\K(2K+1)&-4K-1+3\end{bmatrix} =\begin{bmatrix}-2K+3&4K+2\\K(2K+1)&-4K+2\end{bmatrix}.$$

We must now multiply $$A$$ by this result:

$$A\bigl(A^{3}+3I\bigr)= \begin{bmatrix}0&2\\K&-1\end{bmatrix} \begin{bmatrix}-2K+3&4K+2\\K(2K+1)&-4K+2\end{bmatrix}.$$

Computing each entry one by one,

$$ \begin{aligned} \bigl(A(A^{3}+3I)\bigr)_{11}&= 0\bigl(-2K+3\bigr)+2\cdot K(2K+1)=2K(2K+1),\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{12}&= 0\bigl(4K+2\bigr)+2\bigl(-4K+2\bigr)=-8K+4,\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{21}&= K\bigl(-2K+3\bigr)+(-1)\cdot K(2K+1)=K(-2K+3-2K-1)=-4K^{2}+2K,\\[6pt] \bigl(A(A^{3}+3I)\bigr)_{22}&= K\bigl(4K+2\bigr)+(-1)\bigl(-4K+2\bigr)=4K^{2}+2K+4K-2=4K^{2}+6K-2. \end{aligned} $$

Therefore

$$A(A^{3}+3I)= \begin{bmatrix} 2K(2K+1)&-8K+4\\[4pt] -4K^{2}+2K&4K^{2}+6K-2 \end{bmatrix}.$$

The given condition states that this matrix equals $$2I=\begin{bmatrix}2&0\\0&2\end{bmatrix}$$. We equate corresponding entries:

$$ \begin{aligned} 2K(2K+1)&=2, \quad\text{(1)}\\[4pt] -8K+4&=0, \quad\text{(2)}\\[4pt] -4K^{2}+2K&=0, \quad\text{(3)}\\[4pt] 4K^{2}+6K-2&=2. \quad\text{(4)} \end{aligned} $$

Equation (2) is simplest: $$-8K+4=0 \;\Rightarrow\; 8K=4 \;\Rightarrow\; K=\dfrac12.$$

We must check that this value also satisfies the remaining three equations.

Substituting $$K=\dfrac12$$ into (1):

$$2\left(\dfrac12\right)\left(2\cdot\dfrac12+1\right) =1\cdot2=2,$$

which matches the right-hand side of (1).

For (3):

$$-4\left(\dfrac12\right)^{2}+2\left(\dfrac12\right) =-4\left(\dfrac14\right)+1 =-1+1=0,$$

so (3) is also satisfied.

Finally, (4):

$$4\left(\dfrac12\right)^{2}+6\left(\dfrac12\right)-2 =4\left(\dfrac14\right)+3-2 =1+3-2=2,$$

which again matches the required value. Thus all four equations are consistent with $$K=\dfrac12$$, and no other value obtained from any single equation satisfies the complete set simultaneously.

Therefore $$K=\dfrac12$$ is the unique solution.

Hence, the correct answer is Option A.

We have the three given matrices

First, notice that

so $$A^T=A^{-1}$$ and $$AA^T=I$$. This orthogonality will greatly simplify our work.

The matrix in the statement is

Because a similarity transform does not change powers, the well-known formula

gives at once

We are asked to study

Using associativity and the identities $$AA^T=I$$ and $$A^T A=I$$ we obtain

Thus the matrix whose inverse is required is simply $$B^{2021}$$. Therefore

Next we find $$B^{-1}$$. For a $$2\times2$$ matrix $$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ with $$ad-bc\neq0$$, the inverse is $$\frac1{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$. Here $$\det B=1\cdot1-0\cdot i=1,$$ so

An even easier way is to see $$B=I+N$$ with $$N=\begin{bmatrix}0&0\\ i&0\end{bmatrix},\qquad N^{2}=0,$$ so $$(I+N)^{-1}=I-N$$, yielding the same inverse.

Write $$B^{-1}=I+N',\; N'=\begin{bmatrix}0&0\\-i&0\end{bmatrix}.$$ Again $$\bigl(N'\bigr)^2=0,$$ so the binomial expansion truncates after the linear term:

Therefore

Comparing with the provided options, this matches Option B.

Hence, the correct answer is Option B.

We start with the given matrix

$$P=\begin{bmatrix}1&0\\\dfrac12&1\end{bmatrix}.$$

First we separate it into the sum of the identity matrix and another matrix that has zeros on its diagonal. The $$2\times2$$ identity matrix is

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

Comparing $$P$$ with $$I$$, we notice that the only entry that differs is the left-bottom entry $$\dfrac12$$. So we write

$$P=I+N,$$

where

$$N=\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix}.$$

Now we examine the powers of $$N$$. We multiply $$N$$ by itself:

$$N^2 =\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} \begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} =\begin{bmatrix} 0\cdot0+0\cdot\dfrac12 & 0\cdot0+0\cdot0\\ \dfrac12\cdot0+0\cdot\dfrac12 & \dfrac12\cdot0+0\cdot0 \end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}=0.$$

So $$N$$ is nilpotent of index 2; that is, $$N^2=0$$ and higher powers are also zero:

$$N^3=N^2\cdot N=0,\qquad N^4=0,\quad\text{and so on.}$$

Because $$N^2=0$$, the binomial expansion of $$(I+N)^m$$ truncates after the second term. The general binomial theorem for matrices tells us

$$(I+N)^m=\displaystyle\sum_{k=0}^{m}\binom{m}{k}I^{\,m-k}N^{k}.$$

But since $$N^{2}=0$$, every term with $$k\ge2$$ is zero. So we keep only the $$k=0$$ and $$k=1$$ terms:

$$(I+N)^m=I+\binom{m}{1}N =I+mN.$$

We need the specific case $$m=50$$, therefore

$$P^{50}=(I+N)^{50}=I+50N.$$

Now we substitute $$N=\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix}$$:

$$50N =50\begin{bmatrix}0&0\\\dfrac12&0\end{bmatrix} =\begin{bmatrix}50\cdot0 & 50\cdot0\\ 50\cdot\dfrac12 & 50\cdot0\end{bmatrix} =\begin{bmatrix}0&0\\25&0\end{bmatrix}.$$

Adding this to the identity matrix gives

$$P^{50}=I+50N =\begin{bmatrix}1&0\\0&1\end{bmatrix} +\begin{bmatrix}0&0\\25&0\end{bmatrix} =\begin{bmatrix}1&0\\25&1\end{bmatrix}.$$

We compare this with the options. It coincides exactly with

$$\begin{bmatrix}1&0\\25&1\end{bmatrix},$$

which is Option A.

Hence, the correct answer is Option A.

We are given the matrix $$A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$$ and it is stated that its inverse can be expressed in the form $$A^{-1} = \alpha I + \beta A$$, where $$\alpha, \beta \in \mathbb{R}$$ and $$I$$ is the $$2 \times 2$$ identity matrix. Our task is to find the value of $$4(\alpha - \beta)$$.

First, we explicitly find $$A^{-1}$$. For a general $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, the formula for the inverse (when the determinant is non-zero) is

$$\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right)^{-1} \;=\; \dfrac{1}{ad-bc}\; \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.$$

For our specific matrix, we have $$a = 1, \; b = 2, \; c = -1, \; d = 4$$. We first compute the determinant:

$$\det(A) \;=\; ad - bc \;=\; (1)(4) - (2)(-1) \;=\; 4 + 2 \;=\; 6.$$

Since the determinant is non-zero, the inverse exists. Substituting these values into the formula gives

$$A^{-1} \;=\; \dfrac{1}{6}\; \begin{bmatrix} 4 & -2 \\ 1 & 1 \end{bmatrix}.$$ That is,

$$A^{-1} \;=\; \begin{bmatrix} \dfrac{4}{6} & -\dfrac{2}{6} \\[4pt] \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix} \;=\; \begin{bmatrix} \dfrac{2}{3} & -\dfrac{1}{3} \\[4pt] \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix}.$$

Now we impose the condition $$A^{-1} = \alpha I + \beta A$$. We first write expressions for $$I$$ and $$A$$:

$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}.$$

Multiplying $$A$$ by the scalar $$\beta$$ and then adding $$\alpha I$$, we obtain

$$\alpha I + \beta A \;=\; \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix} \;=\; \begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix}.$$

This matrix must equal $$A^{-1}$$, namely

$$\begin{bmatrix} \alpha + \beta & 2\beta \\ -\beta & \alpha + 4\beta \end{bmatrix} \;=\; \begin{bmatrix} \dfrac{2}{3} & -\dfrac{1}{3} \\ \dfrac{1}{6} & \dfrac{1}{6} \end{bmatrix}.$$

Because two matrices are equal iff all corresponding entries are equal, we equate the entries one by one:

1. From the (1,1) positions: $$\alpha + \beta \;=\; \dfrac{2}{3}.$$

2. From the (1,2) positions: $$2\beta \;=\; -\dfrac{1}{3}.$$ Solving this gives $$\beta = \dfrac{-\frac{1}{3}}{2} = -\dfrac{1}{6}.$$

3. Using $$\beta = -\dfrac{1}{6}$$ in the equation $$\alpha + \beta = \dfrac{2}{3},$$ we find

$$\alpha = \dfrac{2}{3} - \beta = \dfrac{2}{3} - \left(-\dfrac{1}{6}\right) = \dfrac{2}{3} + \dfrac{1}{6} = \dfrac{4}{6} + \dfrac{1}{6} = \dfrac{5}{6}.$$

We have therefore obtained $$\alpha = \dfrac{5}{6}$$ and $$\beta = -\dfrac{1}{6}$$. The quantity required is $$4(\alpha - \beta)$$, so we compute

$$\alpha - \beta \;=\; \dfrac{5}{6} - \left(-\dfrac{1}{6}\right) \;=\; \dfrac{5}{6} + \dfrac{1}{6} \;=\; \dfrac{6}{6} \;=\; 1.$$

Now multiply by 4:

$$4(\alpha - \beta) = 4 \times 1 = 4.$$

Hence, the correct answer is Option D.

We have been given the integral

$$I_{n,m}= \int_{0}^{1/2}\frac{x^{\,n}}{x^{\,m}-1}\,dx,\qquad n>m,\;n,m\in\mathbb N,$$

and in the present problem the value of the parameter $$m$$ is fixed at $$m=3$$.

The entries of the $$3\times3$$ matrix $$A=[a_{ij}]$$ are defined as

$$a_{ij}= \begin{cases} I_{6+i,\,3}-I_{i+3,\,3}, & i\le j,\\[4pt] 0, & i>j. \end{cases}$$

Because each non-zero element depends only on the row-index $$i$$ and because every element below the main diagonal is zero, the matrix $$A$$ is upper-triangular. Consequently the determinant of $$A$$ equals the product of its diagonal elements. To find these diagonal elements we must first evaluate the difference $$I_{6+i,\,3}-I_{i+3,\,3}\;(i=1,2,3).$$

Let us put $$q=i+3\;(i=1,2,3)\;\Longrightarrow\;q=4,5,6$$. Then

$$I_{q+3,\,3}-I_{q,\,3}= \int_{0}^{1/2}\frac{x^{\,q+3}-x^{\,q}}{x^{\,3}-1}\,dx.$$

Observe the algebraic factorisation

$$x^{\,q+3}-x^{\,q}=x^{\,q}(x^{\,3}-1).$$

Substituting this in the integrand we get

$$\frac{x^{\,q+3}-x^{\,q}}{x^{\,3}-1}= \frac{x^{\,q}(x^{\,3}-1)}{x^{\,3}-1}=x^{\,q}.$$

Hence

$$I_{q+3,\,3}-I_{q,\,3}= \int_{0}^{1/2}x^{\,q}\,dx.$$

Using the standard power-rule integral

$$\int x^{\,n}\,dx=\frac{x^{\,n+1}}{n+1}+C,$$

we find

$$I_{q+3,\,3}-I_{q,\,3}= \left[\frac{x^{\,q+1}}{q+1}\right]_{0}^{1/2}= \frac{(1/2)^{\,q+1}}{q+1}\;-\;0= \frac{1}{q+1}\;2^{-(q+1)}.$$

Remembering that $$q=i+3$$, we finally obtain

$$I_{6+i,\,3}-I_{i+3,\,3}= \frac{1}{(i+3)+1}\;2^{-(i+3+1)}=\frac{1}{i+4}\;2^{-(i+4)}.$$

Denote this quantity by $$d_i$$ for ease of writing:

$$d_i=\frac{1}{i+4}\,2^{-(i+4)},\qquad i=1,2,3.$$

Let us list the three required values explicitly:

$$ \begin{aligned} d_1 &=\frac{1}{1+4}\,2^{-(1+4)}=\frac{1}{5}\,2^{-5}= \frac{1}{5\cdot2^{5}}=\frac{1}{160},\\[6pt] d_2 &=\frac{1}{2+4}\,2^{-(2+4)}=\frac{1}{6}\,2^{-6}= \frac{1}{6\cdot2^{6}}=\frac{1}{384},\\[6pt] d_3 &=\frac{1}{3+4}\,2^{-(3+4)}=\frac{1}{7}\,2^{-7}= \frac{1}{7\cdot2^{7}}=\frac{1}{896}. \end{aligned} $$

Since $$A$$ is upper-triangular, its determinant is simply

$$\det A=d_1\,d_2\,d_3.$$

Multiplying the three fractions we get

$$ \begin{aligned} \det A &=\left(\frac{1}{5\cdot2^{5}}\right) \left(\frac{1}{6\cdot2^{6}}\right) \left(\frac{1}{7\cdot2^{7}}\right)\\[6pt] &=\frac{1}{(5\cdot6\cdot7)\;2^{5+6+7}}\\[6pt] &=\frac{1}{210\;2^{18}}. \end{aligned} $$

Clearly $$A$$ is nonsingular, so $$A^{-1}$$ exists. Now recall two standard identities from matrix theory:

1. For any invertible matrix $$B$$, $$\det(B^{-1})=\dfrac{1}{\det B}.$$

2. For any invertible $$n\times n$$ matrix $$B$$, the adjugate satisfies $$\operatorname{adj}B=\det(B)\,B^{-1}$$  and consequently

$$\det(\operatorname{adj}B)=(\det B)^{\,n-1}.$$

We will apply these with $$B=A^{-1}$$ and note that our size is $$n=3$$.

First,

$$\det(A^{-1})=\frac{1}{\det A}=210\;2^{18}.$$

Next, the determinant of the adjugate of $$A^{-1}$$ is, by the second identity,

$$ \det(\operatorname{adj}A^{-1})=\bigl(\det(A^{-1})\bigr)^{\,3-1} =\bigl(210\;2^{18}\bigr)^{2}. $$

Let us expand this square step by step:

$$ \begin{aligned} \bigl(210\;2^{18}\bigr)^{2}&=210^{2}\;(2^{18})^{2}\\[6pt] &=210^{2}\;2^{36}. \end{aligned} $$

Because $$210=2\cdot105,$$ we can factor out a further power of two:

$$ 210^{2}=(2\cdot105)^{2}=2^{2}\cdot105^{2}=4\,105^{2}. $$

Therefore

$$ \det(\operatorname{adj}A^{-1})=4\;105^{2}\;2^{36}=105^{2}\;2^{38}. $$

The numerical expression that matches this result among the options is exactly $$ (105)^{2}\times2^{38}$$, which is listed as Option 4.

Hence, the correct answer is Option 4.

Let $$t = \tan\left(\frac{\theta}{2}\right)$$. Then $$A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$$.

We have $$I_2 + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$ and $$I_2 - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$$.

The determinant of $$I_2 - A$$ is $$1 + t^2$$, so $$(I_2 - A)^{-1} = \frac{1}{1 + t^2}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$$.

Now $$(I_2 + A)(I_2 - A)^{-1} = \frac{1}{1+t^2}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}\begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \frac{1}{1+t^2}\begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2 \end{bmatrix}$$.

Comparing with $$\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$$, we identify $$a = \frac{1-t^2}{1+t^2}$$ and $$b = \frac{2t}{1+t^2}$$.

These are the well-known half-angle identities: $$a = \cos\theta$$ and $$b = \sin\theta$$.

Therefore, $$a^2 + b^2 = \cos^2\theta + \sin^2\theta = 1$$, and $$13(a^2 + b^2) = 13 \times 1 = 13$$.

We are given $$P = \begin{pmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{pmatrix}$$ and $$PQ = kI_3$$ for some non-zero $$k$$. This means $$Q = kP^{-1} = \frac{k}{\det(P)} \text{adj}(P)$$.

We first find $$\det(P)$$. Expanding along the first row, $$\det(P) = 3(0 \cdot 0 - \alpha \cdot (-5)) - (-1)(2 \cdot 0 - \alpha \cdot 3) + (-2)(2 \cdot (-5) - 0 \cdot 3)$$.

$$= 3(5\alpha) + 1(-3\alpha) + (-2)(-10) = 15\alpha - 3\alpha + 20 = 12\alpha + 20$$.

Now we use the condition $$q_{23} = -\frac{k}{8}$$. Since $$Q = \frac{k}{\det(P)} \text{adj}(P)$$, we have $$q_{23} = \frac{k}{\det(P)} \cdot C_{32}$$, where $$C_{32}$$ is the cofactor of the $$(3, 2)$$ entry of $$P$$.

$$C_{32} = (-1)^{3+2} \begin{vmatrix} 3 & -2 \\ 2 & \alpha \end{vmatrix} = -(3\alpha + 4)$$.

So $$q_{23} = \frac{k \cdot (-(3\alpha + 4))}{12\alpha + 20} = -\frac{k}{8}$$.

This gives $$\frac{3\alpha + 4}{12\alpha + 20} = \frac{1}{8}$$. Cross-multiplying, $$8(3\alpha + 4) = 12\alpha + 20$$, so $$24\alpha + 32 = 12\alpha + 20$$, giving $$12\alpha = -12$$ and $$\alpha = -1$$.

Substituting $$\alpha = -1$$, we get $$\det(P) = 12(-1) + 20 = 8$$.

Now we use the condition $$|Q| = \frac{k^2}{2}$$. From $$PQ = kI$$, taking determinants, $$\det(P) \cdot \det(Q) = k^3$$, so $$\det(Q) = \frac{k^3}{\det(P)} = \frac{k^3}{8}$$.

Setting this equal to $$\frac{k^2}{2}$$, we get $$\frac{k^3}{8} = \frac{k^2}{2}$$. Since $$k \neq 0$$, dividing both sides by $$k^2$$ gives $$\frac{k}{8} = \frac{1}{2}$$, so $$k = 4$$.

Therefore, $$\alpha^2 + k^2 = (-1)^2 + 4^2 = 1 + 16 = 17$$.

Hence, the answer is $$17$$.

Write $$A = I + N$$ where $$N = A - I = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$. Since $$N$$ is strictly upper triangular, it is nilpotent: $$N^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ and $$N^3 = 0$$.

By the binomial theorem for commuting matrices, $$A^n = (I+N)^n = I + nN + \binom{n}{2}N^2$$ (since all higher powers of $$N$$ vanish).

The $$(1,3)$$ entry (top-right) of $$I$$ is 0, of $$N$$ is 0, and of $$N^2$$ is 1. So the $$(1,3)$$ entry of $$A^n$$ is $$\binom{n}{2} = \frac{n(n-1)}{2}$$.

Computing: the $$(1,3)$$ entry of $$A^{20}$$ is $$\binom{20}{2} = 190$$, and of $$A^7$$ is $$\binom{7}{2} = 21$$. The $$(1,3)$$ entry of $$I$$ is 0.

Therefore: $$b_{13} = 7 \cdot 190 - 20 \cdot 21 + 2 \cdot 0 = 1330 - 420 = 910$$

We have $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ and $$B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ such that $$AB = B$$.

Writing this out: $$\begin{bmatrix} a\alpha + b\beta \\ c\alpha + d\beta \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$. This gives us $$a\alpha + b\beta = \alpha$$ and $$c\alpha + d\beta = \beta$$, i.e., $$(a-1)\alpha + b\beta = 0$$ and $$c\alpha + (d-1)\beta = 0$$.

Since $$B \neq 0$$, this homogeneous system has a non-trivial solution, which means the determinant of the coefficient matrix must be zero: $$\det\begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix} = 0$$.

Expanding: $$(a-1)(d-1) - bc = 0$$, which gives $$ad - a - d + 1 - bc = 0$$, so $$ad - bc = a + d - 1$$.

Since $$a + d = 2021$$, we get $$ad - bc = 2021 - 1 = 2020$$.

The matrix $$A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$$ is a circulant matrix. Since $$A^2 = I_3$$, we have $$A^2 = I$$, meaning $$A$$ is an involutory matrix, so its eigenvalues are $$\pm 1$$.

For a circulant matrix with first row $$(x, y, z)$$, the eigenvalues are $$\lambda_k = x + y\omega^k + z\omega^{2k}$$ where $$\omega = e^{2\pi i/3}$$ for $$k = 0, 1, 2$$.

The eigenvalue for $$k = 0$$ is $$\lambda_0 = x + y + z$$. Since $$x + y + z > 0$$ and $$\lambda_0 = \pm 1$$, we must have $$x + y + z = 1$$.

Since $$A^2 = I_3$$, we also know $$\det(A)^2 = 1$$, so $$\det(A) = \pm 1$$. The determinant of this circulant matrix is $$x^3 + y^3 + z^3 - 3xyz$$. Since $$xyz = 2$$, we get $$\det(A) = x^3 + y^3 + z^3 - 6$$.

Now using the identity $$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$, we have $$\det(A) = 1 \cdot (x^2 + y^2 + z^2 - xy - yz - zx)$$.

Also, $$A^2 = I$$ means $$AA = I$$. Computing the $$(1,1)$$ entry of $$A^2$$: $$x^2 + y^2 + z^2 = 1$$. Computing the $$(1,2)$$ entry: $$xy + yz + zx = 0$$.

Therefore, $$\det(A) = 1 - 0 = 1$$, and $$x^3 + y^3 + z^3 - 6 = 1$$, giving $$x^3 + y^3 + z^3 = 7$$.

We have $$P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$$. First compute $$P^2 = P \cdot P = \begin{bmatrix} 4-5 & -2+3 \\ 10-15 & -5+9 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$$. Notice that $$P^2 = I - P$$, since $$-P + I = \begin{bmatrix} -2+1 & 1 \\ -5 & 3+1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$$.

Using the recurrence $$P^2 = I - P$$, we compute successive powers. $$P^3 = P \cdot P^2 = P(I - P) = P - P^2 = P - (I - P) = 2P - I$$. Then $$P^4 = P \cdot P^3 = P(2P - I) = 2P^2 - P = 2(I-P) - P = 2I - 3P$$. Continuing: $$P^5 = P \cdot P^4 = P(2I-3P) = 2P - 3P^2 = 2P - 3(I-P) = 5P - 3I$$. Finally, $$P^6 = P \cdot P^5 = P(5P-3I) = 5P^2 - 3P = 5(I-P) - 3P = 5I - 8P$$.

Therefore $$P^n = 5I - 8P$$ when $$n = 6$$.

We need to find the number of $$3 \times 3$$ matrices $$M$$ with entries from $$\{0, 1, 2\}$$ such that the sum of diagonal elements of $$M^T M$$ is seven.

The trace of $$M^T M$$ equals the sum of squares of all entries of $$M$$. This is because $$(M^T M)_{ii} = \sum_j M_{ji}^2$$, so $$\text{tr}(M^T M) = \sum_{i,j} M_{ji}^2$$.

Each entry of $$M$$ is $$0$$, $$1$$, or $$2$$, contributing $$0$$, $$1$$, or $$4$$ to the sum of squares, respectively. We need the total sum of squares of all $$9$$ entries to equal $$7$$.

We consider the possible cases.

Case 1: One entry is $$2$$ (contributing $$4$$), three entries are $$1$$ (contributing $$3$$), and five entries are $$0$$. The sum of squares is $$4 + 3 = 7$$. The number of such matrices is $$\binom{9}{1} \times \binom{8}{3} = 9 \times 56 = 504$$.

Case 2: No entry is $$2$$, so seven entries are $$1$$ (contributing $$7$$) and two entries are $$0$$. The number of such matrices is $$\binom{9}{2} = 36$$.

No other combinations of $$0$$, $$1$$, and $$4$$ can sum to $$7$$ (for instance, two entries equal to $$2$$ would already contribute $$8 > 7$$).

The total number of matrices is $$504 + 36 = 540$$.

Hence, the answer is $$540$$.

We have the given matrix

$$A=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}.$$

First we separate the diagonal (identity) part from the strictly upper-triangular part. Writing $$I$$ for the identity matrix, we observe

$$A=I+U,\qquad\text{where}\qquad U=A-I=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}.$$

Because every entry of $$U$$ that lies on or below the main diagonal is zero, $$U$$ is nilpotent. Let us check the powers of $$U$$ explicitly:

$$U^2 =\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} \begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} =\begin{bmatrix}0\cdot0+1\cdot0+1\cdot0 & 0\cdot1+1\cdot0+1\cdot0 & 0\cdot1+1\cdot1+1\cdot0\\ 0&0&0\\ 0&0&0 \end{bmatrix} =\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},$$

and

$$U^3 =U^2U =\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} \begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix} =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0.$$

Thus $$U^3=0,$$ so $$U$$ has nilpotency index $$3$$.

For any positive integer $$n$$ we need $$A^n=(I+U)^n.$$ Because $$I$$ commutes with $$U,$$ the usual binomial theorem for commuting matrices applies:

$$ (I+U)^n =\sum_{k=0}^{n}\binom{n}{k}I^{\,n-k}U^{\,k} =I+\binom{n}{1}U+\binom{n}{2}U^{2}+\binom{n}{3}U^{3}. $$

Since $$U^{3}=0,$$ all terms with $$k\ge 3$$ vanish. Therefore

$$A^n=I+nU+\frac{n(n-1)}{2}\,U^{2}.$$

Now we form the required matrix

$$M=A+A^{2}+A^{3}+\dots+A^{20}=\sum_{n=1}^{20}A^{n}.$$

Substituting the above expression for $$A^n$$ we obtain

$$ M=\sum_{n=1}^{20}\Bigl(I+nU+\frac{n(n-1)}{2}U^{2}\Bigr) =\Bigl(\sum_{n=1}^{20}I\Bigr) +\Bigl(\sum_{n=1}^{20}n\Bigr)U +\Bigl(\sum_{n=1}^{20}\frac{n(n-1)}{2}\Bigr)U^{2}. $$

Let us evaluate each scalar sum one by one.

The first sum is simply

$$\sum_{n=1}^{20}I = 20I.$$

For the second sum we use the formula for the sum of the first $$N$$ natural numbers, $$1+2+\dots+N=\dfrac{N(N+1)}{2},$$ with $$N=20$$:

$$\sum_{n=1}^{20}n=\frac{20\cdot21}{2}=210.$$

For the third sum we need

$$\sum_{n=1}^{20}\frac{n(n-1)}{2} =\frac12\sum_{n=1}^{20}(n^{2}-n).$$

We compute the two parts separately.

The formula for the sum of squares is $$1^{2}+2^{2}+\dots+N^{2}=\dfrac{N(N+1)(2N+1)}{6}.$$ Putting $$N=20$$ we get

$$\sum_{n=1}^{20}n^{2}=\frac{20\cdot21\cdot41}{6}=2870.$$

We already know $$\sum_{n=1}^{20}n=210.$$ Therefore

$$ \sum_{n=1}^{20}(n^{2}-n)=2870-210=2660, $$

and hence

$$ \sum_{n=1}^{20}\frac{n(n-1)}{2}=\frac{2660}{2}=1330. $$

Putting these scalar sums back, we have

$$M=20I+210\,U+1330\,U^{2}.$$

To get the required answer, we must add all the entries of $$M$$. Because summation of matrix elements is a linear operation, we can add the totals contributed by $$I,\;U$$ and $$U^{2}$$ separately.

The sum of the elements of the identity matrix is clearly

$$\text{Sum}(I)=1+1+1=3.$$

The matrix $$U$$ is

$$U=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix},$$

so

$$\text{Sum}(U)=0+1+1+0+0+1+0+0+0=3.$$

The matrix $$U^{2}$$ is

$$U^{2}=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},$$

thus

$$\text{Sum}(U^{2})=0+0+1+0+0+0+0+0+0=1.$$

Consequently, the total sum of all elements of $$M$$ equals

$$ \begin{aligned} \text{Sum}(M) &=20\cdot\text{Sum}(I)+210\cdot\text{Sum}(U)+1330\cdot\text{Sum}(U^{2})\\ &=20\cdot3+210\cdot3+1330\cdot1\\ &=60+630+1330\\ &=2020. \end{aligned} $$

So, the answer is $$2020$$.

The matrix $$A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is a permutation matrix that swaps rows 1 and 2 when it left-multiplies a matrix, and swaps columns 1 and 2 when it right-multiplies a matrix.

Let $$B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$. Then $$AB$$ is obtained by swapping rows 1 and 2 of $$B$$: $$AB = \begin{bmatrix} b_{21} & b_{22} & b_{23} \\ b_{11} & b_{12} & b_{13} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$.

And $$BA$$ is obtained by swapping columns 1 and 2 of $$B$$: $$BA = \begin{bmatrix} b_{12} & b_{11} & b_{13} \\ b_{22} & b_{21} & b_{23} \\ b_{32} & b_{31} & b_{33} \end{bmatrix}$$.

Setting $$AB = BA$$, we compare entry by entry:

From position $$(1,1)$$: $$b_{21} = b_{12}$$. From $$(1,2)$$: $$b_{22} = b_{11}$$. From $$(1,3)$$: $$b_{23} = b_{13}$$.

From $$(2,1)$$: $$b_{11} = b_{22}$$ (same as above). From $$(2,2)$$: $$b_{12} = b_{21}$$ (same as above). From $$(2,3)$$: $$b_{13} = b_{23}$$ (same as above).

From $$(3,1)$$: $$b_{31} = b_{32}$$. From $$(3,2)$$: $$b_{32} = b_{31}$$ (same). From $$(3,3)$$: $$b_{33} = b_{33}$$ (always true).

So the independent constraints are: $$b_{11} = b_{22}$$, $$b_{12} = b_{21}$$, $$b_{13} = b_{23}$$, and $$b_{31} = b_{32}$$.

The free parameters are: $$b_{11}$$ (which determines $$b_{22}$$), $$b_{12}$$ (which determines $$b_{21}$$), $$b_{13}$$ (which determines $$b_{23}$$), $$b_{31}$$ (which determines $$b_{32}$$), and $$b_{33}$$. That gives 5 free parameters, each taking values from $$\{1, 2, 3, 4, 5\}$$.

The total number of such matrices is $$5^5 = 3125$$.

We have $$A = XB$$ where $$X = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix}$$, giving us $$a_1 = \frac{b_1 - b_2}{\sqrt{3}}$$ and $$a_2 = \frac{b_1 + kb_2}{\sqrt{3}}$$.

Computing $$a_1^2 + a_2^2$$: $$a_1^2 = \frac{(b_1-b_2)^2}{3} = \frac{b_1^2 - 2b_1b_2 + b_2^2}{3}$$ and $$a_2^2 = \frac{(b_1+kb_2)^2}{3} = \frac{b_1^2 + 2kb_1b_2 + k^2b_2^2}{3}$$.

Adding: $$a_1^2 + a_2^2 = \frac{2b_1^2 + (1+k^2)b_2^2 + 2(k-1)b_1b_2}{3}$$.

Setting equal to $$\frac{2}{3}(b_1^2 + b_2^2) = \frac{2b_1^2 + 2b_2^2}{3}$$, and clearing the denominator of 3: $$2b_1^2 + (1+k^2)b_2^2 + 2(k-1)b_1b_2 = 2b_1^2 + 2b_2^2$$.

Simplifying: $$(k^2 - 1)b_2^2 + 2(k-1)b_1b_2 = 0$$, which factors as $$(k-1)\bigl[(k+1)b_2^2 + 2b_1b_2\bigr] = 0$$.

So either $$k = 1$$ or $$(k+1)b_2^2 + 2b_1b_2 = 0$$. We are given the condition $$(k^2+1)b_2^2 \neq -2b_1b_2$$, i.e., $$(k^2+1)b_2^2 + 2b_1b_2 \neq 0$$. Suppose the second factor holds: $$(k+1)b_2^2 + 2b_1b_2 = 0$$, meaning $$2b_1b_2 = -(k+1)b_2^2$$. Substituting into $$(k^2+1)b_2^2 + 2b_1b_2$$: $$(k^2+1)b_2^2 - (k+1)b_2^2 = (k^2 - k)b_2^2 = k(k-1)b_2^2$$. For this to be non-zero (as given), we need $$k \neq 0$$, $$k \neq 1$$, and $$b_2 \neq 0$$. So in principle the second factor could hold without contradicting the given condition when $$k \neq 0, 1$$. However, the second factor $$(k+1)b_2^2 + 2b_1b_2 = 0$$ imposes a specific relationship between $$b_1$$ and $$b_2$$, namely $$b_1 = -\frac{(k+1)b_2}{2}$$. The problem states the condition must hold for the given $$b_1, b_2$$ without any such restriction, so the equation $$(k-1)\bigl[(k+1)b_2^2 + 2b_1b_2\bigr] = 0$$ must be satisfied universally. The only way to guarantee this for arbitrary $$b_1, b_2$$ satisfying the constraint is $$k = 1$$.

When $$k = 1$$, we can verify: $$X = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$$, so $$a_1^2+a_2^2 = \frac{(b_1-b_2)^2+(b_1+b_2)^2}{3} = \frac{2b_1^2+2b_2^2}{3} = \frac{2}{3}(b_1^2+b_2^2)$$, which matches the given condition for all $$b_1, b_2$$.

The answer is $$1$$.

We need to find the number of matrices $$A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}$$ with $$a, b, d \in \{-1, 0, 1\}$$ satisfying $$(I - A)^3 = I - A^3$$.

Step 1: Simplify the condition.

Since $$I$$ commutes with every matrix, we expand:

$$(I - A)^3 = I - 3A + 3A^2 - A^3$$

Setting this equal to $$I - A^3$$:

$$I - 3A + 3A^2 - A^3 = I - A^3$$

$$-3A + 3A^2 = 0$$

$$A^2 = A$$

So $$A$$ must be idempotent.

Step 2: Compute $$A^2$$ and set up equations.

$$A^2 = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} = \begin{pmatrix} a^2 & ab + bd \\ 0 & d^2 \end{pmatrix}$$

Setting $$A^2 = A$$ gives three equations:

$$(i)\; a^2 = a, \quad (ii)\; d^2 = d, \quad (iii)\; b(a + d) = b$$

Step 3: Solve for $$a$$ and $$d$$.

From $$a^2 = a$$ with $$a \in \{-1, 0, 1\}$$:

$$a = -1 \implies 1 \neq -1 \quad (\text{fails})$$

$$a = 0 \implies 0 = 0 \quad (\text{works})$$

$$a = 1 \implies 1 = 1 \quad (\text{works})$$

So $$a \in \{0, 1\}$$, and similarly $$d \in \{0, 1\}$$.

Step 4: Solve for $$b$$ using equation (iii): $$b(a + d - 1) = 0$$.

Case 1: $$(a, d) = (0, 0)$$. Then $$a + d - 1 = -1 \neq 0$$, so $$b = 0$$. This gives 1 matrix.

Case 2: $$(a, d) = (1, 0)$$. Then $$a + d - 1 = 0$$, so $$b$$ can be any of $$\{-1, 0, 1\}$$. This gives 3 matrices.

Case 3: $$(a, d) = (0, 1)$$. Then $$a + d - 1 = 0$$, so $$b \in \{-1, 0, 1\}$$. This gives 3 matrices.

Case 4: $$(a, d) = (1, 1)$$. Then $$a + d - 1 = 1 \neq 0$$, so $$b = 0$$. This gives 1 matrix.

Total count: $$1 + 3 + 3 + 1 = 8$$.

For a $$3 \times 3$$ matrix $$A$$ with entries from $$\{0, 1, 2, 3\}$$, the $$(i,i)$$-th entry of $$AA^T$$ is the sum of squares of the entries in the $$i$$-th row. Specifically, if $$A = (a_{ij})$$, then $$(AA^T)_{ii} = \sum_{j=1}^{3} a_{ij}^2$$.

The sum of all diagonal entries of $$AA^T$$ is $$\sum_{i=1}^{3}\sum_{j=1}^{3} a_{ij}^2 = 9$$. We need the sum of squares of all 9 entries to equal 9.

Each entry $$a_{ij} \in \{0, 1, 2, 3\}$$ contributes $$a_{ij}^2 \in \{0, 1, 4, 9\}$$. We need the total sum of these 9 squared values to be 9.

Since each squared value is at least 0, and the maximum single contribution from entry value 3 is 9, the possible distributions of entry values across 9 positions (where the sum of squares = 9) are:

Case 1: Exactly 9 entries equal to 1 (and 0 entries of other values). Sum of squares = $$9 \times 1 = 9$$. Number of matrices = 1.

Case 2: Exactly one entry equals 2 (contributing 4), exactly 5 entries equal 1 (contributing 5), and 3 entries equal 0. Sum = $$4 + 5 = 9$$. Number of ways = $$\binom{9}{1} \times \binom{8}{5} = 9 \times 56 = 504$$.

Case 3: Exactly two entries equal 2 (contributing 8), exactly 1 entry equals 1 (contributing 1), and 6 entries equal 0. Sum = $$8 + 1 = 9$$. Number of ways = $$\binom{9}{2} \times \binom{7}{1} = 36 \times 7 = 252$$.

Case 4: Exactly one entry equals 3 (contributing 9), and 8 entries equal 0. Sum = 9. Number of ways = $$\binom{9}{1} = 9$$.

Total = $$1 + 504 + 252 + 9 = 766$$.

We find the eigenvalues of $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$$. The characteristic polynomial is $$\det(A - \lambda I) = (2 - \lambda)[(1 - \lambda)(-1 - \lambda)] = -(2 - \lambda)(\lambda^2 - 1)$$, giving eigenvalues $$\lambda_1 = 1$$, $$\lambda_2 = 2$$, and $$\lambda_3 = -1$$.

Since all three eigenvalues are distinct, $$A$$ is diagonalizable, and for any polynomial $$f$$, the eigenvalues of $$f(A)$$ are $$f(\lambda_1)$$, $$f(\lambda_2)$$, and $$f(\lambda_3)$$.

Let $$f(\lambda) = \lambda^{20} + \alpha \lambda^{19} + \beta \lambda$$. The right-hand side matrix has eigenvalues $$1, 4, 1$$. We match each eigenvalue of $$f(A)$$ to the corresponding eigenvalue of the RHS.

From $$f(1) = 1$$: we get $$1 + \alpha + \beta = 1$$, so $$\alpha + \beta = 0$$, meaning $$\beta = -\alpha$$.

From $$f(-1) = 1$$: we get $$1 - \alpha - \beta = 1$$, which gives $$\alpha + \beta = 0$$. This is consistent with the equation above.

From $$f(2) = 4$$: we get $$2^{20} + \alpha \cdot 2^{19} + 2\beta = 4$$. Substituting $$\beta = -\alpha$$ gives $$2^{20} + 2^{19}\alpha - 2\alpha = 4$$, so $$\alpha(2^{19} - 2) = 4 - 2^{20}$$.

Factoring the right side: $$4 - 2^{20} = -4(2^{18} - 1)$$, and the left coefficient: $$2^{19} - 2 = 2(2^{18} - 1)$$. Dividing gives $$\alpha = \frac{-4(2^{18} - 1)}{2(2^{18} - 1)} = -2$$.

Therefore $$\beta = -\alpha = 2$$. We verify: $$f(2) = 2^{20} - 2 \cdot 2^{19} + 2 \cdot 2 = 2^{20} - 2^{20} + 4 = 4$$, which confirms the result.

The answer is $$\beta - \alpha = 2 - (-2) = 4$$.

We work with a $$3\times 3$$ real matrix $$A$$. Throughout, we shall recall three standard facts about determinants and adjugates:

1. For any scalar $$k$$ and matrix $$M$$ of order $$3$$, $$\operatorname{Adj}(kM)=k^{3-1}\operatorname{Adj}(M)=k^{2}\operatorname{Adj}(M).$$

2. For any invertible matrix $$M$$ of order $$3$$, $$\det(\operatorname{Adj}(M))=\det(M)^{3-1}=\det(M)^{2}.$$ (This remains true even if $$M$$ is singular, by continuity.)

3. For order $$3$$, $$\operatorname{Adj}(\operatorname{Adj}(M))=\det(M)^{3-2}\,M=\det(M)\,M.$$

We now unravel the given expression step by step. Put $$B=2A$$ so that $$\det(B)=\det(2A)=2^{3}\det(A)=8\det(A).$$

First inner adjugate:
$$\operatorname{Adj}(B)=\operatorname{Adj}(2A)=2^{2}\operatorname{Adj}(A)=4\,\operatorname{Adj}(A).$$

Next, applying fact 3 to $$B$$:

$$\operatorname{Adj}(\operatorname{Adj}(B))=\det(B)\,B=\det(2A)\,(2A)=8\det(A)\,(2A)=16\det(A)\,A.$$

Multiply this by the scalar $$2$$ that sits immediately outside it:

$$D=2\,\operatorname{Adj}(\operatorname{Adj}(B))=2\,(16\det(A)\,A)=32\det(A)\,A.$$

We must now take the adjugate of $$D$$ and afterwards multiply once more by the leading scalar $$2$$ present in the original problem. Using fact 1 on $$D=kA$$ with $$k=32\det(A)\,,$$ we have

$$\operatorname{Adj}(D)=\operatorname{Adj}(32\det(A)\,A)=(32\det(A))^{2}\operatorname{Adj}(A)=1024\,\det(A)^{2}\operatorname{Adj}(A).$$

Placing the outermost factor $$2$$ finally gives the matrix whose determinant is prescribed:

$$M=2\,\operatorname{Adj}(D)=2\left(1024\,\det(A)^{2}\operatorname{Adj}(A)\right)=2048\,\det(A)^{2}\operatorname{Adj}(A)=32\,\det(2A)^{2}\operatorname{Adj}(A).$$

Observe that $$2048=2^{11}$$ and, crucially, $$\det(2A)=8\det(A)$$; so in determinant form we write succinctly

$$M=32\,\bigl(\det(2A)\bigr)^{2}\operatorname{Adj}(A).$$

We now evaluate $$\det(M)$$. Because $$M$$ is a product of a scalar matrix factor and $$\operatorname{Adj}(A)$$, we use $$\det(kM)=k^{3}\det(M)$$:

$$ \det(M) =\bigl(32\,\det(2A)^{2}\bigr)^{3}\,\det\!\bigl(\operatorname{Adj}(A)\bigr). $$

Insert the separate determinants one by one:

• $$32=2^{5} \Longrightarrow 32^{3}=2^{15}.$$br> • $$\det(2A)^{2}=(8\det(A))^{2}=64\det(A)^{2} \Longrightarrow (\,\det(2A)^{2}\,)^{3}=64^{3}\det(A)^{6}=2^{18}\det(A)^{6}.$$br> • By fact 2, $$\det\!\bigl(\operatorname{Adj}(A)\bigr)=\det(A)^{2}.$$

Putting all these together,

$$ \det(M)=2^{15}\;\times\;2^{18}\,\det(A)^{6}\;\times\;\det(A)^{2} =2^{33}\,\det(A)^{8}. $$

The problem states $$\det(M)=2^{41}$$, so we equate:

$$2^{33}\,\det(A)^{8}=2^{41}\quad\Longrightarrow\quad \det(A)^{8}=2^{41-33}=2^{8}.$$

Taking the real eighth root gives $$\det(A)=2$$(the sign is immaterial here, because we soon square the determinant).

Finally, to reach the required quantity, recall $$\det(A^{2})=\bigl(\det(A)\bigr)^{2}$$ for any square matrix. Hence

$$\det(A^{2})=\Bigl(2\Bigr)^{2}=4.$$

So, the answer is $$4$$.

We have a general matrix from the set

$$A=\begin{pmatrix}a & b\\c & d\end{pmatrix},\qquad a,b,c,d\in\{-3,-2,-1,0,1,2,3\}.$$

For any such matrix the determinant is defined by the well-known formula

$$\det A = ad-bc.$$

In this question we must count how many ordered quadruples $$(a,b,c,d)$$ satisfy the condition

$$ad-bc = 15.$$

First concentrate on the term $$ad$$. With the given entries the largest possible product is $$3\cdot 3 = 9$$ and the smallest is $$(-3)\cdot 3=-9$$. Hence

$$-9\le ad\le 9.$$

Rewrite the determinant condition as

$$bc = ad-15.$$ So the required product $$bc$$ is $$bc = t,\qquad\text{where }t = ad-15.$$

Because $$ad\in[-9,9]$$, the possible values of $$t$$ lie between

$$-9-15=-24\quad\text{and}\quad 9-15=-6,$$ that is, $$-24\le t\le -6.$$

But $$bc$$ itself is the product of two elements from the same set $$\{-3,-2,-1,0,1,2,3\}$$, and such a product can only take values in the interval $$[-9,9]$$. Therefore the overlap of the two ranges is

$$-9\le t\le -6.$$

Thus $$t$$ can be $$-9,-8,-7,-6$$. We next decide which of these actually occur.

Because $$t=ad-15$$, we solve for $$ad$$

$$ad = t+15.$$ Substituting each possible $$t$$ gives $$$ \begin{aligned} t=-9 &\;\Longrightarrow\; ad = 6,\\ t=-8 &\;\Longrightarrow\; ad = 7,\\ t=-7 &\;\Longrightarrow\; ad = 8,\\ t=-6 &\;\Longrightarrow\; ad = 9. \end{aligned} $$$

Remember that $$ad$$ itself must be attainable with $$a,d\in\{-3,-2,-1,0,1,2,3\}$$. Let us list all achievable products and their corresponding ordered pairs:

$$\bullet\;ad=6$$: possible through $$2\cdot3$$, $$3\cdot2$$, $$(-2)\cdot(-3)$$, $$(-3)\cdot(-2)$$. That gives the four ordered pairs $$(a,d)=(2,3),\;(3,2),\;(-2,-3),\;(-3,-2).$$

$$\bullet\;ad=7$$: impossible because neither $$7$$ nor $$-7$$ can be written as a product of two numbers from the allowed set.

$$\bullet\;ad=8$$: impossible for the same reason; $$8$$ has factors $$2$$ and $$4$$ or $$-2$$ and $$-4$$, but $$\pm4\notin\{-3,-2,-1,0,1,2,3\}$$.

$$\bullet\;ad=9$$: possible through $$3\cdot3$$ and $$(-3)\cdot(-3)$$, giving the two ordered pairs $$(a,d)=(3,3),\;(-3,-3).$$

Hence only $$ad=6$$ or $$ad=9$$ are feasible. Their corresponding required values of $$bc$$ are

$$$ \begin{aligned} ad=6 &\;\Longrightarrow\; bc = 6-15 = -9,\\ ad=9 &\;\Longrightarrow\; bc = 9-15 = -6. \end{aligned} $$$

Now we count the ordered pairs $$(b,c)$$ that give each product.

$$\bullet\;bc=-9$$: because $$3\cdot3=9$$ we need opposite signs. The pairs are $$(b,c)=(3,-3),\;(-3,3).$$ So there are $$2$$ possibilities.

$$\bullet\;bc=-6$$: here $$2\cdot3=6$$. Again the signs must be opposite, yielding $$(b,c)=(2,-3),\;(-2,3),\;(3,-2),\;(-3,2).$$ This gives $$4$$ possibilities.

Finally form the matrices by combining the independent choices for $$(a,d)$$ and $$(b,c)$$.

$$$ \begin{aligned} ad=6 &\text{ (4 choices)} \quad\&\quad bc=-9 &\text{ (2 choices)} &\;\Longrightarrow\; 4\times2 = 8\text{ matrices},\\ ad=9 &\text{ (2 choices)} \quad\&\quad bc=-6 &\text{ (4 choices)} &\;\Longrightarrow\; 2\times4 = 8\text{ matrices}. \end{aligned} $$$

Adding the two disjoint cases we obtain the total

$$8+8 = 16.$$

So, the answer is $$16$$.

We are given the matrix $$A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}$$ and the identity matrix $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$. Our task is to find $$10A^{-1}$$ and then compare it with the four expressions provided in the options.

First, we recall the standard formula for the inverse of a $$2 \times 2$$ matrix. For a general matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, the inverse exists (provided the determinant is non-zero) and is given by

$$\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)^{-1} \;=\; \dfrac{1}{ad-bc}\,\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$

Here $$a = 2,\; b = 2,\; c = 9,\; d = 4$$. So we first compute the determinant:

$$\det(A) \;=\; ad - bc \;=\; (2)(4) - (2)(9) \;=\; 8 - 18 \;=\; -10.$$

Since the determinant is $$-10 \neq 0$$, the inverse exists. Using the same formula, the adjugate (or adjoint) matrix of $$A$$ is

$$\operatorname{adj}(A) \;=\; \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \;=\; \begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}.$$

Therefore, the inverse of $$A$$ is

$$A^{-1} \;=\; \dfrac{1}{\det(A)}\,\operatorname{adj}(A) \;=\; \dfrac{1}{-10}\,\begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}.$$

Carrying the scalar $$\dfrac{1}{-10}$$ inside, we get

$$A^{-1} \;=\; \begin{pmatrix} -\dfrac{4}{10} & \dfrac{2}{10} \\[4pt] \dfrac{9}{10} & -\dfrac{2}{10} \end{pmatrix}.$$ By simplifying the individual fractions we may keep them as they are, or proceed directly to the next requirement, which is to multiply the whole inverse by $$10$$.

Multiplying each entry of $$A^{-1}$$ by $$10$$ gives us

$$10A^{-1} \;=\; 10 \times \begin{pmatrix} -\dfrac{4}{10} & \dfrac{2}{10} \\[4pt] \dfrac{9}{10} & -\dfrac{2}{10} \end{pmatrix} \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}.$$

We now compare this matrix with the four candidate expressions. Let us explicitly evaluate each option.

We first compute $$A - 6I$$ because the numeric difference $$6$$ directly appears in two of the options and will likely match the magnitude of the numbers we obtained.

Since $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$, we have $$6I = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}.$$ Therefore,

$$A - 6I \;=\; \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix} \;-\; \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix} \;=\; \begin{pmatrix} 2-6 & 2-0 \\ 9-0 & 4-6 \end{pmatrix} \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}.$$

We see that

$$A - 6I \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix} \;=\; 10A^{-1}.$$

Hence, the matrix $$10A^{-1}$$ is exactly equal to $$A - 6I$$. This corresponds to Option C in the list.

Hence, the correct answer is Option C.

We have the quadratic equation $$x^{2}+x+1=0$$. Since the constant term and the coefficient of the middle term are both $$1$$, its roots are the non-real cube roots of unity. Hence any root $$\alpha$$ of this equation satisfies the two standard relations

$$\alpha^{2}+\alpha+1=0 \quad\Longrightarrow\quad \alpha^{3}=1,\qquad\text{and therefore}\qquad\alpha^{4}=\alpha.$$

The given matrix is written more compactly by extracting the common scalar factor:

$$A=\dfrac1{\sqrt3}\,M,\quad\text{where}\quad M=\begin{bmatrix} 1&1&1\\ 1&\alpha&\alpha^{2}\\ 1&\alpha^{2}&\alpha^{4} \end{bmatrix} =\begin{bmatrix} 1&1&1\\ 1&\alpha&\alpha^{2}\\ 1&\alpha^{2}&\alpha \end{bmatrix}.$$

To obtain powers of $$A$$ we first compute $$M^{2}$$ entry by entry. The general rule used is $$(M^{2})_{ij}=R_{i}\cdot C_{j},$$ where $$R_{i}$$ is the $$i^{\text{th}}$$ row and $$C_{j}$$ the $$j^{\text{th}}$$ column of $$M$$.

First row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{11}&=1+1+1=3,\\ (M^{2})_{12}&=1+\alpha+\alpha^{2}=0,\\ (M^{2})_{13}&=1+\alpha^{2}+\alpha=0. \end{aligned}$$

Second row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{21}&=1+\alpha+\alpha^{2}=0,\\ (M^{2})_{22}&=1+\alpha^{2}+\alpha=0,\\ (M^{2})_{23}&=1+\alpha\alpha^{2}+\alpha^{2}\alpha=1+1+1=3. \end{aligned}$$

Third row of $$M^{2}$$

$$\begin{aligned} (M^{2})_{31}&=1+\alpha^{2}+\alpha=0,\\ (M^{2})_{32}&=1+\alpha^{2}\alpha+\alpha\alpha^{2}=1+1+1=3,\\ (M^{2})_{33}&=1+\alpha+\alpha^{2}=0. \end{aligned}$$

Collecting all entries gives

$$M^{2}= \begin{bmatrix} 3&0&0\\ 0&0&3\\ 0&3&0 \end{bmatrix}.$$

Using $$A=\dfrac1{\sqrt3}M$$ we now square $$A$$:

$$A^{2}=\left(\dfrac1{\sqrt3}M\right)^{2} =\dfrac1{3}\,M^{2} =\dfrac1{3}\begin{bmatrix} 3&0&0\\ 0&0&3\\ 0&3&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix}.$$

The matrix obtained is merely a permutation matrix that keeps the first component intact and swaps the second and third components. Denote it by $$P$$, viz.

$$P=\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&1&0 \end{bmatrix},\qquad\text{so that}\qquad A^{2}=P.$$

A permutation matrix squared either remains the same or returns to the identity; here, swapping twice restores the original order, so

$$P^{2}=I_{3}\quad\Longrightarrow\quad A^{4}=(A^{2})^{2}=P^{2}=I_{3}.$$

The equality $$A^{4}=I_{3}$$ shows that the order of $$A$$ divides $$4$$. To find $$A^{31}$$ we reduce the exponent modulo $$4$$:

$$31=4\times7+3\;\Longrightarrow\;A^{31}=A^{(4\times7)+3}=(A^{4})^{7}\,A^{3}=I_{3}^{\,7}\,A^{3}=A^{3}.$$

Therefore $$A^{31}=A^{3}$$, which matches Option A.

Hence, the correct answer is Option A.

We need to find all $$2 \times 2$$ matrices $$A$$ with entries from $$\{0, 1\}$$ such that $$|A| \neq 0$$, and then check statements (P) and (Q).

Let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ where $$a, b, c, d \in \{0, 1\}$$ and $$|A| = ad - bc \neq 0$$.

Since entries are 0 or 1, the determinant $$ad - bc$$ can only be $$-1$$, $$0$$, or $$1$$. We need $$|A| \neq 0$$, so $$|A| = 1$$ or $$|A| = -1$$.

Case 1: $$|A| = 1$$ (i.e., $$ad = 1$$ and $$bc = 0$$)

$$ad = 1$$ requires $$a = 1$$ and $$d = 1$$. $$bc = 0$$ requires at least one of $$b, c$$ to be 0.

The matrices are:

$$A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2$$, with $$\text{tr}(A_1) = 2$$

$$A_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$, with $$\text{tr}(A_2) = 2$$

$$A_3 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$, with $$\text{tr}(A_3) = 2$$

Case 2: $$|A| = -1$$ (i.e., $$ad = 0$$ and $$bc = 1$$)

$$bc = 1$$ requires $$b = 1$$ and $$c = 1$$. $$ad = 0$$ requires at least one of $$a, d$$ to be 0.

The matrices are:

$$A_4 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$, with $$\text{tr}(A_4) = 0$$

$$A_5 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$, with $$\text{tr}(A_5) = 1$$

$$A_6 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$, with $$\text{tr}(A_6) = 1$$

Now we check each statement.

Statement (P): If $$A \neq I_2$$, then $$|A| = -1$$.

Consider $$A_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$. Here $$A_2 \neq I_2$$, but $$|A_2| = 1 \times 1 - 1 \times 0 = 1 \neq -1$$.

This is a counterexample, so statement (P) is false.

Statement (Q): If $$|A| = 1$$, then $$\text{tr}(A) = 2$$.

From Case 1, all matrices with $$|A| = 1$$ are $$A_1, A_2, A_3$$, and every one of them has $$\text{tr}(A) = 2$$.

So statement (Q) is true.

Therefore, (P) is false and (Q) is true, which corresponds to Option A.

We have the angle $$\theta=\dfrac{\pi}{5}\;(=36^{\circ})$$ and the matrix

$$ A=\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta& \cos\theta \end{bmatrix}. $$

This is the standard rotation matrix, so $$\det(A)=1.$$ A power of a rotation matrix is again a rotation matrix: in fact

$$ A^{n}=\begin{bmatrix} \cos(n\theta) & \sin(n\theta)\\ -\sin(n\theta)& \cos(n\theta) \end{bmatrix} \quad\text{for every integer }n. $$

Taking $$n=4$$ we obtain

$$ A^{4}=\begin{bmatrix} \cos(4\theta) & \sin(4\theta)\\ -\sin(4\theta)& \cos(4\theta) \end{bmatrix}. $$

Now define $$B=A+A^{4}.$$ Adding the two rotation matrices entry-wise we get

$$ B=\begin{bmatrix} \cos\theta+\cos4\theta & \;\sin\theta+\sin4\theta\\ -(\sin\theta+\sin4\theta) & \;\cos\theta+\cos4\theta \end{bmatrix}. $$

For any matrix of the form $$\begin{bmatrix}a & b\\ -b & a\end{bmatrix},$$ the determinant is $$a^{2}+b^{2},$$ because

$$ \det\begin{bmatrix}a & b\\ -b & a\end{bmatrix}=a\cdot a-(-b)\,b=a^{2}+b^{2}. $$

So, with $$a=\cos\theta+\cos4\theta,\qquad b=\sin\theta+\sin4\theta,$$ we have

$$ \det(B)=\bigl(\cos\theta+\cos4\theta\bigr)^{2}+\bigl(\sin\theta+\sin4\theta\bigr)^{2}. $$

We evaluate the two trigonometric sums. First use the sum-to-product identity

$$ \cos x+\cos y=2\cos\!\left(\dfrac{x+y}{2}\right)\cos\!\left(\dfrac{x-y}{2}\right). $$

Putting $$x=\theta,\;y=4\theta$$ gives

$$ \cos\theta+\cos4\theta =2\cos\!\left(\dfrac{\theta+4\theta}{2}\right)\cos\!\left(\dfrac{\theta-4\theta}{2}\right) =2\cos\!\left(\dfrac{5\theta}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right). $$

Because $$5\theta=\pi,$$ we have $$\dfrac{5\theta}{2}=\dfrac{\pi}{2},$$ and $$\cos\dfrac{\pi}{2}=0.$$ Hence

$$ \cos\theta+\cos4\theta=0. $$

Next use the sum-to-product identity for sines,

$$ \sin x+\sin y=2\sin\!\left(\dfrac{x+y}{2}\right)\cos\!\left(\dfrac{x-y}{2}\right), $$

again with $$x=\theta,\;y=4\theta:$$

$$ \sin\theta+\sin4\theta =2\sin\!\left(\dfrac{5\theta}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right) =2\sin\!\left(\dfrac{\pi}{2}\right)\cos\!\left(-\dfrac{3\theta}{2}\right). $$

Since $$\sin\dfrac{\pi}{2}=1$$ and $$\cos(-x)=\cos x,$$ we get

$$ \sin\theta+\sin4\theta =2\cos\!\left(\dfrac{3\theta}{2}\right) =2\cos\!\left(\dfrac{3\pi}{10}\right). $$

Hence

$$ \det(B)=0^{2}+\Bigl(2\cos\dfrac{3\pi}{10}\Bigr)^{2} =4\cos^{2}\dfrac{3\pi}{10}. $$

The exact value of $$\cos\dfrac{3\pi}{10}=\cos54^{\circ}$$ is known:

$$ \cos54^{\circ}=\dfrac{\sqrt{10-2\sqrt5}}{4}. $$

Squaring and multiplying by $$4$$ we obtain

$$ \det(B)=4\left(\dfrac{10-2\sqrt5}{16}\right) =\dfrac{10-2\sqrt5}{4} =\dfrac{5-\sqrt5}{2}. $$

The numerical value is

$$ \dfrac{5-\sqrt5}{2}\approx\dfrac{5-2.236}{2}\approx\dfrac{2.764}{2}\approx1.382. $$

This lies strictly between $$1$$ and $$2.$$ Therefore $$\det(B)$$ belongs to the interval $$(1,2).$$

Hence, the correct answer is Option D.

We are given the adjugate (adjoint) of a $$3 \times 3$$ matrix $$A$$ as

First, we recall the basic relation between a square matrix and its adjugate:

For a matrix of order $$n$$, another very useful fact is

Since we are dealing with a $$3 \times 3$$ matrix, $$n=3$$ and therefore

So, if we compute the determinant of the given $$\operatorname{adj}A$$, call it $$\Delta$$, we can immediately get $$|A|$$ from the equation

Let us now calculate $$\Delta$$ step by step. Using the first row expansion for the determinant, we have

Thus

Setting this equal to $$|A|^{\,2}$$ we get

The problem denotes $$|A|$$ by $$\lambda$$. We shall need only the absolute value, so

Next, we must evaluate $$\mu = \bigl|\,(B^{-1})^{T}\bigr|$$ where

There is a standard identity for the double adjugate (valid for all nonsingular $$n \times n$$ matrices):

With $$n=3$$ this becomes

Let us now find $$B^{-1}$$. Because $$|A|$$ is just a scalar, we have

Taking the transpose, we get

The determinant of a transpose equals the determinant of the original matrix, and extracting a scalar $$\frac{1}{|A|}$$ from a $$3 \times 3$$ determinant raises it to the third power. Hence

But $$|A^{-1}| = \dfrac{1}{|A|}$$, so we obtain

Substituting $$|A| = \pm 3$$ (whose absolute value is $$3$$) gives

Collecting the two required numbers, we have

Hence, the correct answer is Option A.

We are told that the same unknown matrix $$A$$ satisfies three different systems

$$A\,x_1 = b_1,\qquad A\,x_2 = b_2,\qquad A\,x_3 = b_3.$$

We collect the three solution vectors as the columns of one matrix

$$X = \begin{bmatrix} 1 & 0 & 0\\[2pt] 1 & 2 & 0\\[2pt] 1 & 1 & 1 \end{bmatrix},$$

and the corresponding right-hand sides as the columns of another matrix

$$B = \begin{bmatrix} 1 & 0 & 0\\[2pt] 0 & 2 & 0\\[2pt] 0 & 0 & 2 \end{bmatrix}.$$

The three given relations combine into the single matrix equation

$$A\,X = B.$$

If a square matrix multiplies another square matrix on the right and gives a square result, then, provided the middle matrix is invertible, we may isolate the left matrix. The formula we use is

$$\text{If } A\,X = B \text{ and } X \text{ is invertible, then } A = B\,X^{-1}.$$

The determinant of a product equals the product of the determinants, so

$$\det(A) = \dfrac{\det(B)}{\det(X)}.$$

Now we evaluate the two determinants separately.

Determinant of $$B$$

The matrix $$B$$ is diagonal with entries $$1, 2, 2$$ along the diagonal, hence

$$\det(B) = 1 \times 2 \times 2 = 4.$$

Determinant of $$X$$

For the matrix

$$X = \begin{bmatrix} 1 & 0 & 0\\ 1 & 2 & 0\\ 1 & 1 & 1 \end{bmatrix},$$

we expand about the first row:

$$\det(X) = 1 \times \det\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} \;-\; 0 \times (\text{something}) \;+\; 0 \times (\text{something}).$$

The remaining $$2 \times 2$$ determinant is

$$\det\begin{bmatrix} 2 & 0\\ 1 & 1 \end{bmatrix} = 2\cdot1 - 0\cdot1 = 2.$$

So

$$\det(X) = 1 \times 2 = 2.$$

Putting the values together

Substituting into the earlier ratio, we get

$$\det(A) = \dfrac{\det(B)}{\det(X)} = \dfrac{4}{2} = 2.$$

Hence, the correct answer is Option B.

We have the matrix

$$A=\begin{bmatrix}\cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta\end{bmatrix},\qquad\theta=\frac{\pi}{24},\qquad i=\sqrt{-1}.$$

To make the powers of $$A$$ easy to handle, write the $$2\times2$$ identity matrix as $$I$$ and introduce the matrix

$$J=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$

Observe that $$J^2=I$$ because

$$\begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.$$

Now rewrite $$A$$ in terms of $$I$$ and $$J$$:

$$A=\cos\theta\,I+i\sin\theta\,J.$$

Since $$I$$ and $$J$$ commute ($$IJ=JI$$), we may use the standard trigonometric-exponential identity

$$\bigl(\cos\theta\,I+i\sin\theta\,J\bigr)^n=\cos(n\theta)\,I+i\sin(n\theta)\,J,$$

which is the matrix analogue of $$\bigl(\cos\theta+i\sin\theta\bigr)^n=\cos(n\theta)+i\sin(n\theta)$$ (De Moivre’s formula).

Putting $$n=5$$ we obtain

$$A^5=\cos(5\theta)\,I+i\sin(5\theta)\,J =\begin{bmatrix} \cos(5\theta) & i\sin(5\theta)\\ i\sin(5\theta) & \cos(5\theta) \end{bmatrix}.$$

Comparing with $$A^5=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ gives

$$a=d=\cos(5\theta),\qquad b=c=i\sin(5\theta).$$

Square each entry:

$$a^2=d^2=\cos^2(5\theta),$$

$$b^2=c^2=(i\sin(5\theta))^2=i^2\sin^2(5\theta)=-\sin^2(5\theta).$$

Because $$\theta=\dfrac{\pi}{24}$$,

$$5\theta=\frac{5\pi}{24},\qquad 10\theta=\frac{10\pi}{24}=\frac{5\pi}{12}=75^{\circ}.$$

With these preparations we can check each option.

Option A:

$$a^2+b^2=\cos^2(5\theta)+\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)-\sin^2(5\theta) =\cos\bigl(2\cdot5\theta\bigr) =\cos(10\theta) =\cos75^{\circ}\approx0.2588.$$

This value satisfies $$0\le a^2+b^2\le1$$, so Option A is true.

Option B:

$$a^2-d^2=\cos^2(5\theta)-\cos^2(5\theta)=0,$$

so Option B is true.

Option C:

$$a^2-c^2=\cos^2(5\theta)-\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)+\sin^2(5\theta)=1,$$

so Option C is true.

Option D:

$$a^2-b^2=\cos^2(5\theta)-\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)+\sin^2(5\theta)=1,$$

whereas Option D claims this difference equals $$\dfrac12$$. The claim is therefore false.

Exactly one statement must be “not true,” and we have just seen that Option D is the only false one.

Hence, the correct answer is Option D.

We begin with the information that the real, non-zero numbers $$a,\;b,\;c$$ are placed in the matrix

$$A=\begin{bmatrix}a&b&c\\[2pt]b&c&a\\[2pt]c&a&b\end{bmatrix}$$

and that this matrix satisfies $$A^{T}A=I_{3}$$. The columns of an orthogonal matrix are orthonormal, so the dot-product of a column with itself equals $$1$$ and the dot-product of two different columns equals $$0$$. Writing out these conditions gives

$$\begin{aligned} &(1)\quad a^{2}+b^{2}+c^{2}=1,\\[4pt] &(2)\quad ab+bc+ca=0. \end{aligned}$$

We shall denote the elementary symmetric sums by

$$S_{1}=a+b+c,\qquad S_{2}=ab+bc+ca,\qquad S_{3}=abc.$$

From (2) we already have $$S_{2}=0$$. Squaring $$S_{1}$$ and using (1) and (2) we get

$$S_{1}^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=1+2\cdot0=1,$$

so

$$S_{1}=a+b+c=\pm1.\qquad(3)$$

Next we employ the well-known identity

$$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)\bigl(a^{2}+b^{2}+c^{2}-ab-bc-ca\bigr).$$

Because of (1) and (2) the factor in parentheses is simply $$1$$, therefore

$$a^{3}+b^{3}+c^{3}-3abc=a+b+c=S_{1}.$$

The question tells us that $$a^{3}+b^{3}+c^{3}=2$$, so substituting this value we obtain the linear relation

$$2-3S_{3}=S_{1}.\qquad(4)$$

Using (3) inside (4) we analyse the two possible signs separately.

• If $$S_{1}=1$$ then $$2-3S_{3}=1\Longrightarrow3S_{3}=1\Longrightarrow S_{3}=abc=\dfrac13.$$

• If $$S_{1}=-1$$ then $$2-3S_{3}=-1\Longrightarrow3S_{3}=3\Longrightarrow S_{3}=abc=1.$$

So the algebra allows two numerical values for $$abc$$, namely $$\dfrac13$$ and $$1$$.

However, the four options offered in the problem statement are

$$-\dfrac13,\;\dfrac13,\;3,\;\dfrac23,$$

among which only $$\dfrac13$$ appears. Consequently, that is the admissible value demanded by the question.

Hence, the correct answer is Option B.

We begin by recalling a standard fact from linear algebra: for any real matrix $$A$$, the trace of the matrix product $$AA^T$$ is equal to the sum of the squares of all the entries of $$A$$. Symbolically, we write

$$\operatorname{tr}(AA^T)=\sum_{i=1}^{3}\sum_{j=1}^{3}a_{ij}^2,$$

where $$A=(a_{ij})$$ is our $$3\times 3$$ matrix. This happens because the diagonal entry in the $$i$$-th row and $$i$$-th column of $$AA^T$$ is $$\sum_{j=1}^{3}a_{ij}^2$$, and adding the three diagonal entries together simply gathers every $$a_{ij}^2$$ once.

Now the question demands that the sum of the diagonal elements of $$AA^T$$ be $$3$$. Using the equality just stated, we translate that requirement into a very concrete arithmetic condition:

$$\sum_{i=1}^{3}\sum_{j=1}^{3}a_{ij}^2=3.$$

Next, we look at the allowed values of each entry $$a_{ij}$$. Every entry must come from the set $$\{-1,0,1\}$$. We immediately notice

$$(-1)^2 = 1,\qquad 0^2 = 0,\qquad 1^2 = 1.$$

So each individual square is either $$0$$ or $$1$$, and never anything else. Therefore, the total sum of all nine squares can only equal the count of how many entries are non-zero. To make that sum equal to $$3$$, we must have

• exactly three entries of $$A$$ equal to either $$1$$ or $$-1$$ (since each such entry contributes $$1$$ to the sum), and
• the remaining six entries equal to $$0$$ (since $$0^2=0$$ contributes nothing).

Thus, our task converts to a purely combinatorial counting problem: choose any three positions in the $$3\times3$$ grid to hold the non-zero numbers, and then decide independently for each chosen position whether the non-zero number is $$1$$ or $$-1$$.

First, we count the ways to choose the positions. There are $$9$$ positions altogether, and we want any $$3$$ of them, so by the binomial coefficient formula we have

$$\binom{9}{3}= \frac{9\times8\times7}{3\times2\times1}=84.$$

Second, after fixing those three positions, each one can carry either $$1$$ or $$-1$$. The two choices are independent for the three spots, giving in total

$$2^3 = 8$$

sign assignments.

Finally, by the multiplication principle of counting, we multiply the two independent counts:

$$\binom{9}{3}\times 2^3 = 84 \times 8 = 672.$$

Therefore, there are exactly $$672$$ matrices satisfying the given condition.

So, the answer is $$672$$.

We have the matrix $$$A=\begin{bmatrix}x & 1\\ 1 & 0\end{bmatrix}\,.$$$

First we find $$A^2$$. Using the rule “to multiply two matrices, we take the dot-product of rows with columns”, we obtain

$$$A^2=A\cdot A=\begin{bmatrix}x & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix}x & 1\\ 1 & 0\end{bmatrix}= \begin{bmatrix} x\cdot x+1\cdot 1 & x\cdot 1+1\cdot 0\\ 1\cdot x+0\cdot 1 & 1\cdot 1+0\cdot 0 \end{bmatrix} =\begin{bmatrix} x^{2}+1 & x\\ x & 1 \end{bmatrix}.$$$

Now we calculate $$A^3=A^2\cdot A$$:

$$$A^3=\begin{bmatrix}x^{2}+1 & x\\ x & 1\end{bmatrix} \begin{bmatrix}x & 1\\ 1 & 0\end{bmatrix} =\begin{bmatrix} (x^{2}+1)x+x\cdot 1 & (x^{2}+1)\cdot 1 + x\cdot 0\\[4pt] x\cdot x +1\cdot 1 & x\cdot 1 +1\cdot 0 \end{bmatrix} =\begin{bmatrix} x^{3}+2x & x^{2}+1\\ x^{2}+1 & x \end{bmatrix}.$$$

Next we form $$A^4=A^3\cdot A$$:

$$$A^4=\begin{bmatrix}x^{3}+2x & x^{2}+1\\ x^{2}+1 & x\end{bmatrix} \begin{bmatrix}x & 1\\ 1 & 0\end{bmatrix} =\begin{bmatrix} (x^{3}+2x)x+(x^{2}+1)\cdot1 & (x^{3}+2x)\cdot1+(x^{2}+1)\cdot0\\[4pt] (x^{2}+1)x+x\cdot1 & (x^{2}+1)\cdot1+x\cdot0 \end{bmatrix} =\begin{bmatrix} x^{4}+3x^{2}+1 & x^{3}+2x\\ x^{3}+2x & x^{2}+1 \end{bmatrix}.$$$

Thus the $$a_{11}$$ entry of $$A^4$$ is $$x^{4}+3x^{2}+1$$. We are told that $$a_{11}=109$$, so

$$$x^{4}+3x^{2}+1=109 \;\Longrightarrow\; x^{4}+3x^{2}-108=0.$$$

Letting $$y=x^{2}\;(y\ge 0)$$ converts this quartic into the quadratic

$$y^{2}+3y-108=0.$$

Using the quadratic-formula $$y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\;b=3,\;c=-108$$, we find

$$$y=\dfrac{-3\pm\sqrt{3^{2}-4(1)(-108)}}{2}=\dfrac{-3\pm\sqrt{9+432}}{2}=\dfrac{-3\pm21}{2}.$$$

This gives $$y=\dfrac{18}{2}=9$$ or $$y=\dfrac{-24}{2}=-12$$. Because $$y=x^{2}\ge0$$, we discard the negative root and keep $$x^{2}=9$$, i.e. $$x=\pm3$$.

The $$a_{22}$$ entry of $$A^4$$ is, from our earlier result, $$x^{2}+1$$. Substituting $$x^{2}=9$$ yields

$$a_{22}=9+1=10.$$

So, the answer is $$10$$.

We are given that the matrix $$B=\begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix}$$ is the inverse of a matrix $$A$$, that is, $$A^{-1}=B$$.

For any square matrices, we know the basic determinant relation $$\det(A)\,\det(A^{-1}) = 1$$. Since $$A^{-1}=B$$, this becomes

$$\det(A)\,\det(B)=1.$$

The condition given in the question is $$\det(A)+1=0$$. Re-writing, we have

$$\det(A) = -1.$$

Substituting this value of $$\det(A)$$ into the product relation, we get

$$(-1)\,\det(B) = 1 \quad\Longrightarrow\quad \det(B) = -1.$$

So our task reduces to finding all values of $$\alpha$$ for which the determinant of the matrix $$B$$ equals $$-1$$.

Now we compute $$\det(B)$$. Expanding along the first row, we obtain

$$ \det(B) = 5\, \begin{vmatrix} 2 & 1\\ 3 & -1 \end{vmatrix} \;-\; (2\alpha)\, \begin{vmatrix} 0 & 1\\ \alpha & -1 \end{vmatrix} \;+\; 1\, \begin{vmatrix} 0 & 2\\ \alpha & 3 \end{vmatrix}. $$

We evaluate each of the 2×2 determinants one by one.

First minor:

$$ \begin{vmatrix} 2 & 1\\ 3 & -1 \end{vmatrix} = 2(-1) - 1(3) = -2 - 3 = -5. $$

Second minor:

$$ \begin{vmatrix} 0 & 1\\ \alpha & -1 \end{vmatrix} = 0(-1) - 1(\alpha) = -\alpha. $$

Third minor:

$$ \begin{vmatrix} 0 & 2\\ \alpha & 3 \end{vmatrix} = 0\cdot 3 - 2\alpha = -2\alpha. $$

Substituting these into the expansion of $$\det(B)$$, we get

$$ \det(B) \;=\; 5(-5)\;-\;(2\alpha)(-\alpha)\;+\;1(-2\alpha). $$

Carrying out the multiplications:

$$ \det(B) = -25 + 2\alpha^{2} - 2\alpha. $$

We must have $$\det(B) = -1$$, so we set

$$ -25 + 2\alpha^{2} - 2\alpha = -1. $$

Adding 25 to both sides and simplifying,

$$ 2\alpha^{2} - 2\alpha - 24 = 0. $$

Dividing the entire equation by 2,

$$ \alpha^{2} - \alpha - 12 = 0. $$

This is a quadratic equation. Using the quadratic formula $$\alpha = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\; b=-1,\; c=-12$$, we find

$$ \alpha = \frac{1 \pm \sqrt{(-1)^{2}-4(1)(-12)}}{2} = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm \sqrt{49}}{2} = \frac{1 \pm 7}{2}. $$

Thus we get two values:

$$ \alpha_{1} = \frac{1+7}{2} = \frac{8}{2} = 4,\quad \alpha_{2} = \frac{1-7}{2} = \frac{-6}{2} = -3. $$

The question asks for the sum of all such values, so

$$ \alpha_{1} + \alpha_{2} = 4 + (-3) = 1. $$

Hence, the correct answer is Option B.

We have to evaluate the product

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix}$$

and we are told that the result equals

$$\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}.$$

First, recall the general multiplication rule for 2 × 2 matrices. If we have two matrices of the special upper-triangular form $$A=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}, \qquad B=\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix},$$ then their product is obtained as follows:

$$AB=\begin{bmatrix} 1\cdot 1+ a\cdot 0 & 1\cdot b + a\cdot 1\\ 0\cdot 1+1\cdot 0 & 0\cdot b + 1\cdot 1 \end{bmatrix} =\begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix}.$$

So, multiplying two such matrices simply adds their upper-right entries. Because of associativity of matrix multiplication, we can apply this fact successively. Hence, the product of all the given matrices is

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 1+2+3+\dots+(n-1) \\ 0 & 1 \end{bmatrix}.$$

The sum in the upper-right corner is the sum of the first $$n-1$$ natural numbers. The well-known formula for this sum is

$$1+2+3+\dots+(n-1)=\frac{(n-1)n}{2}.$$

Therefore the product becomes

$$\begin{bmatrix} 1 & \dfrac{(n-1)n}{2} \\ 0 & 1 \end{bmatrix}.$$

We are told that this equals $$\begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}.$$ So the upper-right entries must be equal, giving the equation

$$\frac{(n-1)n}{2}=78.$$

Multiplying both sides by 2, we get

$$(n-1)n=156.$$

Expanding and arranging in standard quadratic form,

$$n^2-n-156=0.$$

To solve this quadratic, compute the discriminant: $$\Delta = (-1)^2 - 4(1)(-156)=1+624=625.$$ Taking the square root, $$\sqrt{625}=25$$.

Using the quadratic formula $$n=\dfrac{-b\pm\sqrt{\Delta}}{2a}$$ with $$a=1$$ and $$b=-1$$, we get

$$n=\frac{\,1\pm25\,}{2}.$$

This gives two roots: $$n=\frac{26}{2}=13 \quad\text{or}\quad n=\frac{-24}{2}=-12.$$

The problem concerns positive integers, so we accept

$$n=13.$$

Now we must find the inverse of the matrix $$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 13 \\ 0 & 1 \end{bmatrix}.$$

For any matrix of the form $$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix},$$ the inverse is obtained by changing the sign of $$a$$, because

$$\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -a \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & a-a \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$

which is the identity matrix. Therefore, the inverse of $$\begin{bmatrix} 1 & 13 \\ 0 & 1 \end{bmatrix}$$ is

$$\begin{bmatrix} 1 & -13 \\ 0 & 1 \end{bmatrix}.$$

Hence, the correct answer is Option D.

We are told that $$A$$ is a symmetric matrix and $$B$$ is a skew-symmetric matrix, and together they satisfy the relation

$$A+B=\begin{pmatrix}2&3\\5&-1\end{pmatrix}.$$

First, we recall the defining properties of the two kinds of matrices involved:

• For a symmetric matrix we have the formula $$A^T=A$$, which, in the $$2\times2$$ case, forces the off-diagonal entries to be equal.

• For a skew-symmetric matrix we have the formula $$B^T=-B$$, which, in the $$2\times2$$ case, forces the diagonal entries to be zero and the off-diagonal entries to be negatives of each other.

So we write the most general forms respecting these facts. Let

$$A=\begin{pmatrix}a&c\\c&d\end{pmatrix},\qquad B=\begin{pmatrix}0&x\\-x&0\end{pmatrix}.$$

Adding the two matrices, we have

$$A+B=\begin{pmatrix}a&c+x\\c-x&d\end{pmatrix}.$$

The question gives this sum explicitly, so we equate corresponding entries:

$$\begin{aligned} a &= 2,\\ c+x &= 3,\\ c-x &= 5,\\ d &= -1. \end{aligned}$$

Now we solve for the unknowns $$c$$ and $$x$$. Adding the second and third equations eliminates $$x$$:

$$ (c+x)+(c-x)=3+5 \;\Longrightarrow\; 2c=8 \;\Longrightarrow\; c=4. $$

Substituting $$c=4$$ into $$c+x=3$$, we obtain

$$ 4 + x = 3 \;\Longrightarrow\; x = -1. $$

With all the variables known, the individual matrices are

$$A=\begin{pmatrix}2&4\\4&-1\end{pmatrix},\qquad B=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

Our objective is the product $$AB$$. Using standard matrix multiplication, whose formula is $$\bigl(AB\bigr)_{ij}=\sum_{k=1}^{2}A_{ik}B_{kj},$$ we compute each entry one by one.

• First-row, first-column entry:

$$ (AB)_{11}=A_{11}B_{11}+A_{12}B_{21}=2\cdot0+4\cdot1=4. $$

• First-row, second-column entry:

$$ (AB)_{12}=A_{11}B_{12}+A_{12}B_{22}=2\cdot(-1)+4\cdot0=-2. $$

• Second-row, first-column entry:

$$ (AB)_{21}=A_{21}B_{11}+A_{22}B_{21}=4\cdot0+(-1)\cdot1=-1. $$

• Second-row, second-column entry:

$$ (AB)_{22}=A_{21}B_{12}+A_{22}B_{22}=4\cdot(-1)+(-1)\cdot0=-4. $$

Collecting these results, we find

$$AB=\begin{pmatrix}4&-2\\-1&-4\end{pmatrix}.$$

This matches exactly with Option C.

Hence, the correct answer is Option C.

We recall a basic property of determinants: for any two square matrices of the same order we have the formula $$\det(AB)=\det(A)\,\det(B)$$. We also know that $$\det(A^T)=\det(A)$$ and $$\det(A^{-1})=\dfrac{1}{\det(A)}$$. All matrices here are of order $$3 \times 3$$, so these facts apply directly.

First, we are told that

$$\det(ABA^T)=8.$$

Using the determinant product rule and the fact that a transpose does not change a determinant, we write

$$\det(ABA^T)=\det(A)\,\det(B)\,\det(A^T)=\det(A)\,\det(B)\,\det(A).$$

Simplifying, this becomes

$$\det(ABA^T)=\bigl(\det(A)\bigr)^2\,\det(B)=8.$$

Next, we are also given

$$\det(AB^{-1})=8.$$

Again applying the product rule and the inverse‐determinant relation, we have

$$\det(AB^{-1})=\det(A)\,\det(B^{-1})=\det(A)\,\dfrac{1}{\det(B)}=8.$$

To keep the algebra clear, let us denote

$$x=\det(A), \qquad y=\det(B).$$

The two pieces of information now translate to the pair of equations

$$x^{2}y=8 \qquad (1)$$

and

$$\dfrac{x}{y}=8 \qquad (2).$$

From equation (2) we immediately find

$$y=\dfrac{x}{8}.$$

Substituting this value of $$y$$ into equation (1) gives

$$x^{2}\left(\dfrac{x}{8}\right)=8.$$

This simplifies step by step:

$$\dfrac{x^{3}}{8}=8,$$

so

$$x^{3}=64,$$

and therefore

$$x=4.$$

Using $$x=4$$ in the relation $$y=\dfrac{x}{8}$$ we get

$$y=\dfrac{4}{8}=\dfrac{1}{2}.$$

Now we need the determinant

$$\det(BA^{-1}B^{T}).$$

Applying the same determinant properties, we write

$$\det(BA^{-1}B^{T})=\det(B)\,\det(A^{-1})\,\det(B^{T}).$$

Since $$\det(B^{T})=\det(B)$$ and $$\det(A^{-1})=\dfrac{1}{\det(A)}$$, this becomes

$$\det(BA^{-1}B^{T})=y \times \dfrac{1}{x} \times y=\dfrac{y^{2}}{x}.$$

Substituting $$x=4$$ and $$y=\dfrac{1}{2}$$, we calculate

$$\det(BA^{-1}B^{T})=\dfrac{\left(\dfrac{1}{2}\right)^{2}}{4}=\dfrac{\dfrac{1}{4}}{4}=\dfrac{1}{16}.$$

Hence, the correct answer is Option C.

We have the matrix $$A=\begin{pmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{pmatrix}.$$

This is the standard rotation matrix in the plane. A basic fact from linear algebra is that multiplying two rotation matrices adds their angles. Explicitly, for any real numbers $$\theta_1,\,\theta_2$$,

$$\begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1\end{pmatrix}\, \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2\end{pmatrix} =\begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2)\end{pmatrix}.$$

By repeated use of this rule, the power $$A^{n}$$ is still a rotation matrix whose angle is the sum of $$n$$ copies of $$\alpha$$, namely $$n\alpha$$. Therefore,

$$A^{n}=\begin{pmatrix}\cos(n\alpha) & -\sin(n\alpha)\\ \sin(n\alpha) & \cos(n\alpha)\end{pmatrix}.$$

Now we are told that $$A^{32}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}.$$ Comparing this with the general formula just obtained, we can equate the angles:

$$\cos(32\alpha)=0,\quad -\sin(32\alpha)=-1\quad\Longrightarrow\quad \begin{cases} \cos(32\alpha)=0,\\ \sin(32\alpha)=1. \end{cases}$$

Both conditions hold when the angle is $$\dfrac{\pi}{2}$$ plus an integral multiple of $$2\pi$$. Symbolically,

$$32\alpha=\dfrac{\pi}{2}+2k\pi,\qquad k\in\mathbb Z.$$

Solving for $$\alpha$$ gives

$$\alpha=\dfrac{\dfrac{\pi}{2}+2k\pi}{32} =\dfrac{\pi}{64}+\dfrac{k\pi}{16},\qquad k\in\mathbb Z.$$

We now look for those values among the options provided. Setting $$k=0$$ yields

$$\alpha=\dfrac{\pi}{64},$$

which is exactly Option 3 in the list.

Other integral choices of $$k$$ give angles that differ by multiples of $$\dfrac{\pi}{16}$$ and hence do not match the remaining options. Thus, the only option consistent with the given equation is $$\alpha=\dfrac{\pi}{64}$$.

Hence, the correct answer is Option 3.

We start with the rotation matrix

$$A \;=\; \begin{pmatrix}\cos\theta & -\sin\theta\\[4pt] \sin\theta & \cos\theta\end{pmatrix}.$$

A well-known property of rotation matrices is the power formula

$$A^{n}\;=\;\begin{pmatrix}\cos(n\theta) & -\sin(n\theta)\\[4pt] \sin(n\theta) & \cos(n\theta)\end{pmatrix},$$

valid for every integer $$n$$ (positive, negative, or zero). This can be proved by simple mathematical induction or by identifying the matrix with the complex number $$\cos\theta+i\sin\theta$$ and using De Moivre’s theorem.

For a negative exponent we simply write

$$A^{-n}=A^{\,(-n)}=\begin{pmatrix}\cos(-n\theta) & -\sin(-n\theta)\\[4pt] \sin(-n\theta) & \cos(-n\theta)\end{pmatrix}.$$

In the present problem we need $$A^{-50}$$ with $$\theta=\dfrac{\pi}{12}.$$ Substituting the given value of $$\theta$$, we obtain the combined rotation angle:

$$-50\theta \;=\;-50\left(\dfrac{\pi}{12}\right) \;=\;-\dfrac{50\pi}{12} \;=\;-\dfrac{25\pi}{6}.$$

To bring this angle into a more familiar range, we reduce it modulo $$2\pi$$:

$$-\dfrac{25\pi}{6} \;=\;-\left(\dfrac{24\pi}{6}+\dfrac{\pi}{6}\right) \;=\;-4\pi-\dfrac{\pi}{6}.$$

Because a rotation by any integer multiple of $$2\pi$$ leaves the matrix unchanged, the term $$-4\pi$$ can be discarded, leaving

$$-4\pi-\dfrac{\pi}{6}\equiv -\dfrac{\pi}{6}\pmod{2\pi}.$$

Hence

$$A^{-50} =\begin{pmatrix} \cos\!\left(-\dfrac{\pi}{6}\right) & -\sin\!\left(-\dfrac{\pi}{6}\right)\\[8pt] \sin\!\left(-\dfrac{\pi}{6}\right) & \cos\!\left(-\dfrac{\pi}{6}\right) \end{pmatrix}.$$

Next we recall the exact trigonometric values for $$\dfrac{\pi}{6}$$:

$$\cos\!\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt3}{2}, \qquad \sin\!\left(\dfrac{\pi}{6}\right)=\dfrac12.$$ Because cosine is an even function and sine is an odd function, we have

$$\cos\!\left(-\dfrac{\pi}{6}\right)=\cos\!\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt3}{2},$$ $$\sin\!\left(-\dfrac{\pi}{6}\right)=-\sin\!\left(\dfrac{\pi}{6}\right)=-\dfrac12.$$

Substituting these values gives

$$A^{-50} =\begin{pmatrix} \dfrac{\sqrt3}{2} & -\!\left(-\dfrac12\right)\\[8pt] -\dfrac12 & \dfrac{\sqrt3}{2} \end{pmatrix} =\begin{pmatrix} \dfrac{\sqrt3}{2} & \dfrac12\\[8pt] -\dfrac12 & \dfrac{\sqrt3}{2} \end{pmatrix}.$$

On comparing with the options, we observe that this matrix matches Option A exactly.

Hence, the correct answer is Option A.

We are given the real $$3 \times 3$$ matrix

$$A=\begin{pmatrix}0 & 2q & r\\ p & q & -r\\ p & -q & r\end{pmatrix}$$

and the condition $$AA^T=I_3.$$ For any matrix, the relation $$AA^T=I$$ means that every row of $$A$$ has length $$1$$ (unit vectors) and that any two distinct rows are orthogonal (their dot-product is $$0$$). We therefore write the three rows explicitly:

$$R_1=(0,\;2q,\;r),\qquad R_2=(p,\;q,\;-r),\qquad R_3=(p,\;-q,\;r).$$

First, we impose the unit-length requirement. The square of the length of a vector $$(x_1,x_2,x_3)$$ is $$x_1^{\,2}+x_2^{\,2}+x_3^{\,2}$$. Hence

$$\|R_1\|^2=0^2+(2q)^2+r^2=4q^2+r^2=1,$$

$$\|R_2\|^2=p^2+q^2+(-r)^2=p^2+q^2+r^2=1,$$

$$\|R_3\|^2=p^2+(-q)^2+r^2=p^2+q^2+r^2=1.$$

Next, we apply the orthogonality condition. The dot-product of two vectors $$(x_1,x_2,x_3)$$ and $$(y_1,y_2,y_3)$$ is $$x_1y_1+x_2y_2+x_3y_3$$. Therefore

$$R_1\cdot R_2=0\cdot p+(2q)\,q+r\,(-r)=2q^2-r^2=0,$$

$$R_1\cdot R_3=0\cdot p+(2q)\,(-q)+r\,r=-2q^2+r^2=0,$$

$$R_2\cdot R_3=p\cdot p+q\,(-q)+(-r)\,r=p^2-q^2-r^2=0.$$

We now solve these equations step by step.

From $$2q^2-r^2=0$$ we have

$$r^2=2q^2.$$

Substituting this into the length equation $$4q^2+r^2=1$$ gives

$$4q^2+2q^2=1\quad\Longrightarrow\quad6q^2=1\quad\Longrightarrow\quad q^2=\frac16.$$

Using $$r^2=2q^2$$ again, we find

$$r^2=2\left(\frac16\right)=\frac13.$$

Finally, we substitute $$q^2=\dfrac16$$ and $$r^2=\dfrac13$$ into $$p^2+q^2+r^2=1$$:

$$p^2+\frac16+\frac13=1\quad\Longrightarrow\quad p^2+\frac16+\frac26=1\quad\Longrightarrow\quad p^2+\frac36=1\quad\Longrightarrow\quad p^2+\frac12=1.$$

So

$$p^2=1-\frac12=\frac12,$$

whence

$$|p|=\sqrt{\frac12}=\frac1{\sqrt2}.$$

Hence, the correct answer is Option C.

We have the two given matrices

$$P=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}\qquad\text{and}\qquad Q=[q_{ij}]$$

together with the relation

$$Q-P^{5}=I_{3}.$$

This implies immediately that

$$Q=P^{5}+I_{3}.$$

So our task is to find the fifth power of the matrix $$P$$, add the identity matrix $$I_{3}$$ to it, and then read off the required entries of $$Q$$.

Because $$P$$ is a lower-triangular matrix whose diagonal entries are all $$1$$, every power of $$P$$ will also be lower-triangular with diagonal entries $$1$$. We proceed by repeated multiplication, writing out every step clearly.

First power (already known):

$$P^{1}=P=\begin{bmatrix}1&0&0\\3&1&0\\9&3&1\end{bmatrix}.$$

Second power: we use the definition of matrix multiplication. For the entry in the second row and first column, for example, we multiply the second row of the left matrix by the first column of the right matrix. Carrying this out for every position we get

$$\begin{aligned} P^{2}&=P\cdot P\\ &=\begin{bmatrix} 1\cdot1+0\cdot3+0\cdot9 & 1\cdot0+0\cdot1+0\cdot3 & 1\cdot0+0\cdot0+0\cdot1\\ 3\cdot1+1\cdot3+0\cdot9 & 3\cdot0+1\cdot1+0\cdot3 & 3\cdot0+1\cdot0+0\cdot1\\ 9\cdot1+3\cdot3+1\cdot9 & 9\cdot0+3\cdot1+1\cdot3 & 9\cdot0+3\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 6&1&0\\ 27&6&1 \end{bmatrix}. \end{aligned}$$

Third power: multiply $$P^{2}$$ by $$P$$.

$$\begin{aligned} P^{3}&=P^{2}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 6&1&0\\ 27&6&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1\cdot1+0\cdot3+0\cdot9 & 1\cdot0+0\cdot1+0\cdot3 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 6\cdot1+1\cdot3+0\cdot9 & 6\cdot0+1\cdot1+0\cdot3 & 6\cdot0+1\cdot0+0\cdot1\\[4pt] 27\cdot1+6\cdot3+1\cdot9 & 27\cdot0+6\cdot1+1\cdot3 & 27\cdot0+6\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 9&1&0\\ 54&9&1 \end{bmatrix}. \end{aligned}$$

Fourth power: multiply $$P^{3}$$ by $$P$$ once more.

$$\begin{aligned} P^{4}&=P^{3}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 9&1&0\\ 54&9&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1&0&0\\ 9\cdot1+1\cdot3+0\cdot9 & 9\cdot0+1\cdot1+0\cdot3 & 0\\ 54\cdot1+9\cdot3+1\cdot9 & 54\cdot0+9\cdot1+1\cdot3 & 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 12&1&0\\ 90&12&1 \end{bmatrix}. \end{aligned}$$

Fifth power: one final multiplication of $$P^{4}$$ by $$P$$.

$$\begin{aligned} P^{5}&=P^{4}\cdot P\\ &=\begin{bmatrix} 1&0&0\\ 12&1&0\\ 90&12&1 \end{bmatrix} \begin{bmatrix} 1&0&0\\ 3&1&0\\ 9&3&1 \end{bmatrix}\\ &=\begin{bmatrix} 1&0&0\\ 12\cdot1+1\cdot3+0\cdot9 & 12\cdot0+1\cdot1+0\cdot3 & 0\\ 90\cdot1+12\cdot3+1\cdot9 & 90\cdot0+12\cdot1+1\cdot3 & 1 \end{bmatrix}\\[6pt] &=\begin{bmatrix} 1&0&0\\ 15&1&0\\ 135&15&1 \end{bmatrix}. \end{aligned}$$

Now we add the identity matrix $$I_{3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ to obtain $$Q$$:

$$Q=P^{5}+I_{3} =\begin{bmatrix} 1+1 & 0 & 0\\ 15 & 1+1 & 0\\ 135 & 15 & 1+1 \end{bmatrix} =\begin{bmatrix} 2&0&0\\ 15&2&0\\ 135&15&2 \end{bmatrix}.$$

From this explicit form we read off

$$q_{21}=15,\qquad q_{31}=135,\qquad q_{32}=15.$$

We are asked to compute

$$\frac{q_{21}+q_{31}}{q_{32}}=\frac{15+135}{15}=\frac{150}{15}=10.$$

Hence, the correct answer is Option A.

We are given the matrix $$A = \begin{pmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{pmatrix}$$ and the condition $$A^T A = 3I_3$$.

Step 1: Compute $$A^T A$$

The transpose of $$A$$ is:

$$A^T = \begin{pmatrix} 0 & 2x & 2x \\ 2y & y & -y \\ 1 & -1 & 1 \end{pmatrix}$$

The condition $$A^T A = 3I_3$$ means the columns of $$A$$ are orthogonal and each column has magnitude $$\sqrt{3}$$.

Step 2: Column norms equal $$\sqrt{3}$$

Column 1: $$(0, 2x, 2x)^T$$. Norm squared: $$0 + 4x^2 + 4x^2 = 8x^2 = 3$$

$$x^2 = \frac{3}{8}$$, so $$x = \pm\sqrt{\frac{3}{8}}$$

Column 2: $$(2y, y, -y)^T$$. Norm squared: $$4y^2 + y^2 + y^2 = 6y^2 = 3$$

$$y^2 = \frac{1}{2}$$, so $$y = \pm\frac{1}{\sqrt{2}}$$

Column 3: $$(1, -1, 1)^T$$. Norm squared: $$1 + 1 + 1 = 3$$ ✔

Step 3: Verify orthogonality

Column 1 $$\cdot$$ Column 2: $$0 \cdot 2y + 2x \cdot y + 2x \cdot (-y) = 0 + 2xy - 2xy = 0$$ ✔

Column 1 $$\cdot$$ Column 3: $$0 \cdot 1 + 2x \cdot (-1) + 2x \cdot 1 = -2x + 2x = 0$$ ✔

Column 2 $$\cdot$$ Column 3: $$2y \cdot 1 + y \cdot (-1) + (-y) \cdot 1 = 2y - y - y = 0$$ ✔

All orthogonality conditions are automatically satisfied, regardless of the signs of $$x$$ and $$y$$.

Step 4: Check the constraint $$x \neq y$$

We have $$x = \pm\sqrt{3/8}$$ and $$y = \pm 1/\sqrt{2}$$.

Since $$\sqrt{3/8} = \frac{\sqrt{3}}{2\sqrt{2}} \approx 0.612$$ and $$\frac{1}{\sqrt{2}} \approx 0.707$$, these values are never equal.

So all 4 combinations of signs satisfy $$x \neq y$$.

Step 5: Count the matrices

There are 2 choices for $$x$$ and 2 choices for $$y$$, giving $$2 \times 2 = 4$$ matrices.

The correct answer is Option C: 4.

We have a non-singular $$3 \times 3$$ matrix $$A$$ which satisfies the quadratic matrix equation

$$ (A-3I)(A-5I)=O. $$

First we expand the left side exactly as we would expand two algebraic brackets:

$$ (A-3I)(A-5I)=A^2-5A-3A+15I =A^2-8A+15I. $$

Since the product is the zero matrix $$O$$, we obtain

$$ A^2-8A+15I=O \;\;\Longrightarrow\;\; A^2=8A-15I. $$

Because $$A$$ is non-singular it has an inverse $$A^{-1}$$, so we may right-multiply every term by $$A^{-1}$$ (this is legal for matrices of the same size):

$$ A^2A^{-1}=8AA^{-1}-15IA^{-1}. $$

Now we simplify each product:

$$ A^2A^{-1}=A,\qquad AA^{-1}=I,\qquad IA^{-1}=A^{-1}. $$

Substituting these three simplifications gives

$$ A = 8I - 15A^{-1}. $$

We want an explicit expression for $$A^{-1}$$, so we move the term containing $$A^{-1}$$ to the left:

$$ 15A^{-1}=8I-A \;\;\Longrightarrow\;\; A^{-1}=\dfrac{8I-A}{15}. $$

The statement of the problem tells us that real numbers $$\alpha$$ and $$\beta$$ satisfy

$$ \alpha A+\beta A^{-1}=4I. $$

We now replace $$A^{-1}$$ by the expression just found:

$$ \alpha A+\beta\left(\dfrac{8I-A}{15}\right)=4I. $$

Next we distribute the factor $$\beta/15$$ inside the parentheses:

$$ \alpha A+\dfrac{\beta}{15}\,8I-\dfrac{\beta}{15}\,A =4I. $$

Collecting like terms, we separate the coefficients of $$A$$ and $$I$$:

$$\bigl(\alpha-\dfrac{\beta}{15}\bigr)A+\dfrac{8\beta}{15}I =4I.$$

For two matrices to be equal, corresponding coefficients must be equal. Hence

$$ \alpha-\dfrac{\beta}{15}=0 \quad\text{and}\quad \dfrac{8\beta}{15}=4. $$

From the second equation we solve for $$\beta$$:

$$ \dfrac{8\beta}{15}=4 \;\;\Longrightarrow\;\; 8\beta = 60 \;\;\Longrightarrow\;\; \beta = 7.5 =\dfrac{15}{2}. $$

Using $$\alpha=\dfrac{\beta}{15}$$ from the first equation we get

$$ \alpha = \dfrac{\,7.5\,}{15}=\dfrac12. $$

Finally, we add the two numbers:

$$ \alpha+\beta = \dfrac12 + 7.5 = 8. $$

Hence, the correct answer is Option A.

We have the condition that $$A\cdot\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$$ is a scalar matrix. A scalar matrix can be written as $$kI$$, where $$I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$ and $$k$$ is a real number. So the given information is

$$A\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}=kI.$$

Because the matrix on the right is invertible, we can post-multiply both sides by its inverse to isolate $$A$$. The rule is: if $$XY=Z$$ and $$Y$$ is invertible, then $$X=ZY^{-1}$$. Thus,

$$A=kI\left(\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}\right)^{-1}.$$

Now we need the inverse of the upper-triangular matrix $$\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$$. For a matrix $$\begin{bmatrix}a & b\\0 & d\end{bmatrix}$$ the inverse is $$\begin{bmatrix}\dfrac1a & -\dfrac{b}{ad}\\0 & \dfrac1d\end{bmatrix}$$. Substituting $$a=1,\;b=2,\;d=3$$ we get

$$\left(\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}\right)^{-1}= \begin{bmatrix}1 & -\dfrac{2}{3}\\0 & \dfrac13\end{bmatrix}.$$

Multiplying by the scalar $$k$$ gives

$$A=k\begin{bmatrix}1 & -\dfrac{2}{3}\\0 & \dfrac13\end{bmatrix}= \begin{bmatrix}k & -\dfrac{2k}{3}\\0 & \dfrac{k}{3}\end{bmatrix}.$$

Next, the problem tells us $$|3A|=108$$. First compute $$3A$$: multiplying every entry of $$A$$ by 3 results in

$$3A=\begin{bmatrix}3k & -2k\\0 & k\end{bmatrix}.$$

The determinant of an upper-triangular matrix is the product of its diagonal elements. Therefore

$$|3A|=(3k)(k)=3k^2.$$

We are told that $$|3A|=108$$, so

$$3k^2=108\;\;\Longrightarrow\;\;k^2=36\;\;\Longrightarrow\;\;k=\pm6.$$

Because $$k^2$$ will appear in $$A^2$$, the sign of $$k$$ will not matter. We keep $$k^2=36$$ for further calculation.

Now compute $$A^2=A\cdot A$$. Using

$$A=\begin{bmatrix}k & -\dfrac{2k}{3}\\0 & \dfrac{k}{3}\end{bmatrix},$$

we multiply the matrices entry by entry:

$$A^2=\begin{bmatrix}k & -\dfrac{2k}{3}\\0 & \dfrac{k}{3}\end{bmatrix} \begin{bmatrix}k & -\dfrac{2k}{3}\\0 & \dfrac{k}{3}\end{bmatrix}.$$

Row 1, Column 1: $$k\cdot k+\left(-\dfrac{2k}{3}\right)\cdot 0=k^2.$$

Row 1, Column 2: $$k\left(-\dfrac{2k}{3}\right)+\left(-\dfrac{2k}{3}\right)\left(\dfrac{k}{3}\right)= -\dfrac{2k^2}{3}-\dfrac{2k^2}{9}= -\dfrac{8k^2}{9}.$$

Row 2, Column 1: $$0\cdot k+\dfrac{k}{3}\cdot0=0.$$

Row 2, Column 2: $$0\left(-\dfrac{2k}{3}\right)+\dfrac{k}{3}\cdot\dfrac{k}{3}= \dfrac{k^2}{9}.$$

Putting these four results together,

$$A^2=\begin{bmatrix}k^2 & -\dfrac{8k^2}{9}\\0 & \dfrac{k^2}{9}\end{bmatrix}.$$

Finally substitute $$k^2=36$$:

$$A^2=\begin{bmatrix}36 & -\dfrac{8\cdot36}{9}\\0 & \dfrac{36}{9}\end{bmatrix} =\begin{bmatrix}36 & -32\\0 & 4\end{bmatrix}.$$

This matrix matches Option D. Hence, the correct answer is Option D.

We have the lower-triangular matrix

$$A=\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}$$

and we want the sum of the three numbers that appear in the first column of $$B=A^{20}\,.$$ Let us denote, for every positive integer $$n,$$

$$A^{n}=\begin{bmatrix}a_{11}^{(n)}&a_{12}^{(n)}&a_{13}^{(n)}\\[2pt] a_{21}^{(n)}&a_{22}^{(n)}&a_{23}^{(n)}\\[2pt] a_{31}^{(n)}&a_{32}^{(n)}&a_{33}^{(n)}\end{bmatrix}.$$

The required sum is therefore

$$S_{n}=a_{11}^{(n)}+a_{21}^{(n)}+a_{31}^{(n)}.$$

We now build up the powers step by step so that every algebraic detail is visible and a pattern can be recognised.

Power 1. We clearly have

$$A^{1}=A=\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix},$$

so

$$\bigl(a_{11}^{(1)},\,a_{21}^{(1)},\,a_{31}^{(1)}\bigr)=(1,1,1)$$

and therefore

$$S_{1}=1+1+1=3.$$

Power 2. Using plain matrix multiplication, and writing every scalar product explicitly, we have

$$A^{2}=A\!\cdot\!A=\begin{bmatrix} 1\cdot1+0\cdot1+0\cdot1 & 1\cdot0+0\cdot1+0\cdot1 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 1\cdot1+1\cdot1+0\cdot1 & 1\cdot0+1\cdot1+0\cdot1 & 1\cdot0+1\cdot0+0\cdot1\\[4pt] 1\cdot1+1\cdot1+1\cdot1 & 1\cdot0+1\cdot1+1\cdot1 & 1\cdot0+1\cdot0+1\cdot1 \end{bmatrix}$$

which simplifies to

$$A^{2}=\begin{bmatrix}1&0&0\\2&1&0\\3&2&1\end{bmatrix}.$$

Thus

$$\bigl(a_{11}^{(2)},\,a_{21}^{(2)},\,a_{31}^{(2)}\bigr)=(1,2,3)$$

and

$$S_{2}=1+2+3=6.$$

Power 3. Multiplying $$A^{2}$$ by $$A$$ again, while displaying every dot product,

$$A^{3}=A^{2}A=\begin{bmatrix} 1&0&0\\2&1&0\\3&2&1\end{bmatrix} \begin{bmatrix} 1&0&0\\1&1&0\\1&1&1 \end{bmatrix}$$

gives

$$\begin{aligned} A^{3}&=\begin{bmatrix} 1\cdot1+0\cdot1+0\cdot1 & 1\cdot0+0\cdot1+0\cdot1 & 1\cdot0+0\cdot0+0\cdot1\\[4pt] 2\cdot1+1\cdot1+0\cdot1 & 2\cdot0+1\cdot1+0\cdot1 & 2\cdot0+1\cdot0+0\cdot1\\[4pt] 3\cdot1+2\cdot1+1\cdot1 & 3\cdot0+2\cdot1+1\cdot1 & 3\cdot0+2\cdot0+1\cdot1 \end{bmatrix}\\[6pt] &=\begin{bmatrix}1&0&0\\3&1&0\\6&3&1\end{bmatrix}. \end{aligned}$$

Hence

$$\bigl(a_{11}^{(3)},\,a_{21}^{(3)},\,a_{31}^{(3)}\bigr)=(1,3,6)$$

and

$$S_{3}=1+3+6=10.$$

Writing the sums that we have found so far, we notice the sequence

$$S_{1}=3,\qquad S_{2}=6,\qquad S_{3}=10.$$

The successive differences are $$3$$ and $$4,$$ then $$4$$ and $$?$$ in the next step, so it looks like the pattern is the familiar triangular-number pattern. Indeed, the first three terms can be matched exactly by the well-known formula for the $$k$$-th triangular number, namely

$$T_{k}=\frac{k(k+1)}{2}.$$

Re-indexing appropriately, we suspect that

$$S_{n}=\frac{(n+1)(n+2)}{2}=\binom{n+2}{2}.$$

Formal proof of the pattern by mathematical induction.

Base case $$n=1$$ has already been calculated: $$S_{1}=3=\binom{3}{2}.$$ Now assume for some $$n=k$$ that

$$S_{k}=a_{11}^{(k)}+a_{21}^{(k)}+a_{31}^{(k)}=\binom{k+2}{2}.$$

To pass from $$A^{k}$$ to $$A^{k+1},$$ we right-multiply by $$A.$$ Concerning only the first column, we use the fact that the first column of $$A$$ itself is

$$\begin{bmatrix}1\\1\\1\end{bmatrix},$$

so

$$\begin{bmatrix}a_{11}^{(k+1)}\\ a_{21}^{(k+1)}\\ a_{31}^{(k+1)}\end{bmatrix}=A^{k}\begin{bmatrix}1\\1\\1\end{bmatrix}= \begin{bmatrix} a_{11}^{(k)}\cdot1+a_{12}^{(k)}\cdot1+a_{13}^{(k)}\cdot1\\[4pt] a_{21}^{(k)}\cdot1+a_{22}^{(k)}\cdot1+a_{23}^{(k)}\cdot1\\[4pt] a_{31}^{(k)}\cdot1+a_{32}^{(k)}\cdot1+a_{33}^{(k)}\cdot1 \end{bmatrix}.$$

That is,

$$\begin{aligned} a_{11}^{(k+1)}&=a_{11}^{(k)}+a_{12}^{(k)}+a_{13}^{(k)},\\ a_{21}^{(k+1)}&=a_{21}^{(k)}+a_{22}^{(k)}+a_{23}^{(k)},\\ a_{31}^{(k+1)}&=a_{31}^{(k)}+a_{32}^{(k)}+a_{33}^{(k)}. \end{aligned}$$

Adding these three equalities together we get

$$S_{k+1}=S_{k}+ \bigl(a_{12}^{(k)}+a_{22}^{(k)}+a_{32}^{(k)}\bigr)+ \bigl(a_{13}^{(k)}+a_{23}^{(k)}+a_{33}^{(k)}\bigr).$$

But $$A^{k}$$ is lower triangular with ones on the diagonal, so $$a_{12}^{(k)}=0$$ and $$a_{13}^{(k)}=a_{23}^{(k)}=0.$$ Therefore

$$S_{k+1}=S_{k}+a_{22}^{(k)}+a_{32}^{(k)}+a_{33}^{(k)}.$$

Since every diagonal entry of any power of a unit lower-triangular matrix is $$1,$$ we know $$a_{22}^{(k)}=a_{33}^{(k)}=1.$$ Furthermore, the (3,2) entry is precisely $$k,$$ because we can check directly for $$k=1,2,3$$ and see by induction that one unit is added at each new multiplication. Hence

$$S_{k+1}=S_{k}+1+k+1=S_{k}+k+2.$$

Using the induction hypothesis $$S_{k}=\dfrac{(k+1)(k+2)}{2},$$ we get

$$\begin{aligned} S_{k+1}&=\frac{(k+1)(k+2)}{2}+k+2\\[4pt] &=\frac{(k+1)(k+2)+2(k+2)}{2}\\[4pt] &=\frac{(k+2)(k+3)}{2}\\[4pt] &=\binom{(k+1)+2}{2}. \end{aligned}$$

Thus the induction is complete and the formula

$$S_{n}=\frac{(n+1)(n+2)}{2}=\binom{n+2}{2}$$

holds for every positive integer $$n.$$

Evaluating for $$n=20.$$

Substituting $$n=20$$ in the proven expression, we obtain

$$S_{20}=\frac{(20+1)(20+2)}{2}=\frac{21\cdot22}{2}=231.$$

Therefore, the sum of the elements of the first column of $$B=A^{20}$$ equals $$231.$$

Hence, the correct answer is Option D.

We have two relations between the unknown square matrices $$A$$ and $$B$$ of order $$3$$:

$$A + B = 2B' \qquad\qquad (1)$$

$$3A + 2B = I_3 \qquad\qquad (2)$$

The symbol $$B'$$ denotes the transpose of $$B$$ and $$I_3$$ is the identity matrix of order $$3$$.

From relation (1) isolate $$A$$ in terms of $$B$$ and its transpose:

$$A = 2B' - B \qquad\qquad (3)$$

Substitute the expression (3) for $$A$$ in relation (2):

$$3(2B' - B) + 2B \;=\; I_3$$

$$6B' - 3B + 2B \;=\; I_3$$

$$6B' - B \;=\; I_3 \qquad\qquad (4)$$

Because the transpose of the identity matrix is itself, we may take the transpose of both sides of (4) without changing the right-hand side:

$$(6B' - B)' \;=\; I_3'$$

$$6B - B' \;=\; I_3 \qquad\qquad (5)$$

Now equations (4) and (5) give a pair of linear matrix equations in the two unknowns $$B$$ and $$B'$$. Solve them simultaneously. From (4) express $$B$$:

$$B = 6B' - I_3 \qquad\qquad (6)$$

Insert (6) into (5):

$$6(6B' - I_3) - B' \;=\; I_3$$

$$36B' - 6I_3 - B' \;=\; I_3$$

$$35B' \;=\; 7I_3$$

$$B' \;=\; \dfrac{7}{35}I_3 \;=\; \dfrac{1}{5}I_3 \qquad\qquad (7)$$

Transpose (7) to obtain $$B$$ itself (since the transpose of $$I_3$$ is $$I_3$$):

$$B \;=\; \left(B'\right)' \;=\; \dfrac{1}{5}I_3 \qquad\qquad (8)$$

Return to (3) to determine $$A$$:

$$A = 2B' - B$$

Using (7) and (8):

$$A = 2\!\left(\dfrac{1}{5}I_3\right) - \dfrac{1}{5}I_3 = \dfrac{2}{5}I_3 - \dfrac{1}{5}I_3 = \dfrac{1}{5}I_3 \qquad\qquad (9)$$

Thus both matrices are fixed uniquely:

$$A = \dfrac{1}{5}I_3, \qquad B = \dfrac{1}{5}I_3$$

Now evaluate each given option.

Option A:

$$10A + 5B = 10\!\left(\dfrac{1}{5}I_3\right) + 5\!\left(\dfrac{1}{5}I_3\right) = 2I_3 + I_3 = 3I_3$$

Option B:

$$3A + 6B = 3\!\left(\dfrac{1}{5}I_3\right) + 6\!\left(\dfrac{1}{5}I_3\right) = \dfrac{3}{5}I_3 + \dfrac{6}{5}I_3 = \dfrac{9}{5}I_3 \neq 2I_3$$

Option C:

$$5A + 10B = 5\!\left(\dfrac{1}{5}I_3\right) + 10\!\left(\dfrac{1}{5}I_3\right) = I_3 + 2I_3 = 3I_3 \neq 2I_3$$

Option D:

$$B + 2A = \dfrac{1}{5}I_3 + 2\!\left(\dfrac{1}{5}I_3\right) = \dfrac{1}{5}I_3 + \dfrac{2}{5}I_3 = \dfrac{3}{5}I_3 \neq I_3$$

Only Option A produces the identity relation demanded by the option statement.

Hence, the correct answer is Option A.

We have the matrix

$$A=\begin{pmatrix}2 & -3\\-4 & 1\end{pmatrix}.$$

Our goal is to find $$\text{Adj}\,(3A^{2}+12A).$$ We shall proceed step by step, carrying out every algebraic operation in detail.

First, we need the square of $$A$$. For any two-by-two matrices, ordinary matrix multiplication rules apply:

$$A^{2}=A\cdot A=\begin{pmatrix}2 & -3\\-4 & 1\end{pmatrix} \begin{pmatrix}2 & -3\\-4 & 1\end{pmatrix}.$$

Multiplying the first row of the first matrix with the first column of the second gives

$$2\cdot2+(-3)\cdot(-4)=4+12=16.$$

Multiplying the first row with the second column gives

$$2\cdot(-3)+(-3)\cdot1=-6-3=-9.$$

Multiplying the second row with the first column gives

$$(-4)\cdot2+1\cdot(-4)=-8-4=-12.$$

Multiplying the second row with the second column gives

$$(-4)\cdot(-3)+1\cdot1=12+1=13.$$

So

$$A^{2}=\begin{pmatrix}16 & -9\\-12 & 13\end{pmatrix}.$$

Next, we need $$3A^{2}$$. Multiplying every entry of $$A^{2}$$ by $$3$$ gives

$$3A^{2}=3\begin{pmatrix}16 & -9\\-12 & 13\end{pmatrix} =\begin{pmatrix}48 & -27\\-36 & 39\end{pmatrix}.$$

Similarly, $$12A$$ is obtained by multiplying each entry of $$A$$ by $$12$$:

$$12A=12\begin{pmatrix}2 & -3\\-4 & 1\end{pmatrix} =\begin{pmatrix}24 & -36\\-48 & 12\end{pmatrix}.$$

Now we add these two matrices to obtain $$3A^{2}+12A$$:

$$3A^{2}+12A=\begin{pmatrix}48 & -27\\-36 & 39\end{pmatrix} +\begin{pmatrix}24 & -36\\-48 & 12\end{pmatrix} =\begin{pmatrix}48+24 & -27-36\\-36-48 & 39+12\end{pmatrix} =\begin{pmatrix}72 & -63\\-84 & 51\end{pmatrix}.$$

Let us denote this resulting matrix by $$B$$:

$$B=\begin{pmatrix}72 & -63\\-84 & 51\end{pmatrix}.$$

The next task is to find the adjugate (also called the adjoint) of a 2 × 2 matrix. For any matrix

$$\begin{pmatrix}a & b\\c & d\end{pmatrix},$$

the formula for the adjugate is

$$\text{Adj}\,\begin{pmatrix}a & b\\c & d\end{pmatrix} =\begin{pmatrix}d & -b\\-c & a\end{pmatrix}.$$

Applying this formula to $$B$$, we identify

$$a=72,\quad b=-63,\quad c=-84,\quad d=51.$$

Hence

$$\text{Adj}\,B =\begin{pmatrix}d & -b\\-c & a\end{pmatrix} =\begin{pmatrix}51 & -(-63)\\-(-84) & 72\end{pmatrix} =\begin{pmatrix}51 & 63\\84 & 72\end{pmatrix}.$$

This is exactly option B from the list provided.

Hence, the correct answer is Option 2.

For an invertible square matrix we always begin with the basic relation

$$A \; adj(A)=adj(A)\;A=|A|\,I,$$

where $$|A|$$ is the determinant of $$A$$ and $$I$$ is the identity matrix of the same order. This identity is true for every non-singular matrix and will be used repeatedly.

Taking determinant on both sides of $$A\,adj(A)=|A|\,I$$ we have

$$|A\,adj(A)|=||A|\,I|.$$

The left-hand side gives $$|A|\,|adj(A)|$$ while the right-hand side equals $$|A|^{n}$$ because the determinant of $$|A|I$$ (an $$n\times n$$ scalar matrix) is $$|A|^{n}$$. Hence

$$|A|\,|adj(A)|=|A|^{n}\;\;\Longrightarrow\;\;|adj(A)|=|A|^{\,n-1}.$$

In the present problem $$n=3$$, so

$$|adj(A)|=|A|^{\,2}.\qquad(1)$$

Next, from $$A\,adj(A)=|A|\,I$$ we isolate $$adj(A)$$:

$$adj(A)=|A|\,A^{-1}.\qquad(2)$$

Because $$A$$ is invertible, the matrix $$adj(A)$$ is also invertible, and its inverse is obtained directly from (2):

$$(adj(A))^{-1}=\bigl(|A|\,A^{-1}\bigr)^{-1}=\dfrac{1}{|A|}\,A.\qquad(3)$$

Now we wish to express $$adj(adj(A))$$. For any non-singular matrix $$B$$ we have the key identity

$$B\,adj(B)=|B|\,I.$$

We apply it with $$B=adj(A)$$:

$$adj(A)\;adj\!\bigl(adj(A)\bigr)=|adj(A)|\,I.\qquad(4)$$

Using (1) to replace $$|adj(A)|$$ inside (4) gives

$$adj(A)\;adj\!\bigl(adj(A)\bigr)=|A|^{2}\,I.\qquad(5)$$

Now we substitute $$adj(A)=|A|\,A^{-1}$$ from (2) into (5):

$$|A|\,A^{-1}\;adj\!\bigl(adj(A)\bigr)=|A|^{2}\,I.$$

Dividing by $$|A|$$ on both sides, we obtain

$$A^{-1}\;adj\!\bigl(adj(A)\bigr)=|A|\,I.$$

Multiplying on the left by $$A$$ gives the desired explicit expression:

$$adj\!\bigl(adj(A)\bigr)=|A|\,A.\qquad(6)$$

Equation (6) is crucial for comparing with the four given statements.

Let us examine every option one by one.

Option A: The right-hand side is

$$|A|^{2}\,(adj(A))^{-1}=|A|^{2}\left(\dfrac{1}{|A|}A\right)=|A|\,A,$$

which is exactly the left-hand side $$adj(adj(A))$$ from (6). So Option A is always true.

Option B: The right-hand side is

$$|A|\,(adj(A))^{-1}=|A|\left(\dfrac{1}{|A|}A\right)=A,$$

whereas (6) tells us that $$adj(adj(A))=|A|\,A$$. These two matrices are equal only in the special case $$|A|=1$$; in general they are different. Hence Option B is not always true.

Option C: This is exactly equation (6). Therefore it is always true.

Option D: This is formula (2), a standard identity valid for every invertible matrix. Hence it is always true.

Only Option B fails to hold for an arbitrary invertible $$3\times3$$ matrix.

Hence, the correct answer is Option B.