JEE Matrices & Determinants 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.

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): $$\frac{3}{8}$$.

 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.

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

$$a_{23} = -1$$

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: $$16$$

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 $$M^{-1}$$.

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

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

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

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.

Case P (row and column sums zero, no symmetry condition):

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 $$P\rightarrow 12\;.$$

Case Q (symmetric matrices with column sums zero):

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 $$Q\rightarrow 6\;.$$

Case R (skew-symmetric matrix and a linear system):

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 $$R\rightarrow \text{Infinite}\;.$$

Case S (row sums zero, determinant):

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: $$A^3 + A^T$$.

 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 $$\begin{bmatrix} 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}$$. 

 $$|A - 2I| = 0 \cdot \begin{vmatrix} 0 & 11 \\ 3 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 11 \\ 3 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 6 & 0 \\ 3 & 3 \end{vmatrix}$$, 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.

$$ D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} $$

$$ D = 1(5m - 10) - 1(2m - 5) + 1(4 - 5) $$

$$ D = 5m - 10 - 2m + 5 - 1 = 3m - 6 $$

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 $$A = I_2 - 2MM^T$$, 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 $$M$$

 $$M^TX = 0$$.

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

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 $$\boxed{2}$$.

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 $$A^{-1}$$:

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

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

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

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

System equations:

$$2x - 5y = 20$$

$$3x + my = m$$

Using Cramer's Rule:

$$\Delta = |A| = 2m + 15, \quad \Delta_x = 20m + 5m = 25m, \quad \Delta_y = 2m - 60$$

For negative solutions ($$x < 0, y < 0$$): * Case $$\Delta > 0 \implies m > -7.5$$: Requires $$25m < 0 \implies m < 0$$ and $$2m - 60 < 0 \implies m < 30$$. Intersection: $$(-7.5, 0)$$.

Case $$\Delta < 0 \implies m < -7.5$$: Requires $$25m > 0$$ (Impossible for $$m < -7.5$$).

Interval: $$(a,b) = (-7.5, 0) \implies a = -7.5, b = 0$$.

Integral: $$8 \int_{-7.5}^{0} (2m + 15) \, dm = 8 \left[ m^2 + 15m \right]_{-7.5}^{0} = 8 \left(0 - (56.25 - 112.5)\right) = 8 \times 56.25 = \mathbf{450}$$

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 $$\boxed{10}$$.

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

Now

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

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \end{bmatrix}$$

$$a_{11} + a_{12} + a_{13} = 3$$

$$a_{21} + a_{22} + a_{23} = 3$$

$$a_{31} + a_{32} + a_{33} = 3$$

Now for maximum value of $$\det(A)$$, choosing $$a_{ij} = \begin{cases} 0 & i \neq j \\ 3 & i = j \end{cases}$$

$$\therefore \vert{}A\vert{} = 27$$

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 $$4096-N$$.

Case I: $$ad=bc=0$$

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: $$120+105=225.$$

Case II: $$ad=bc\neq0$$ (all four entries non-zero)

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.

A symmetric matrix of order 3 satisfies $$A = A^T$$, meaning $$a_{ij} = a_{ji}$$.

For a 3×3 symmetric matrix, the independent entries are:

• 3 diagonal entries: $$a_{11}, a_{22}, a_{33}$$
• 3 upper triangular entries: $$a_{12}, a_{13}, a_{23}$$

Total independent entries = 6.

Each entry can take any value from the set $$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$$, which has 10 elements.

Number of symmetric matrices = $$10^6$$.

This matches option 2: $$10^6$$.

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.

Given $$B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$$, $$\alpha > 2$$, is the adjoint of matrix $$A$$ with $$|A| = 2$$.

Finding $$\alpha$$:

Since $$B = \text{adj}(A)$$, we have $$|B| = |A|^{n-1} = 2^2 = 4$$ (for $$n = 3$$).

Computing $$|B|$$:

$$|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha)$$

$$= 8 - 3\alpha - 12 + 9\alpha - \alpha^2 = -\alpha^2 + 6\alpha - 4$$

Setting $$|B| = 4$$:

$$-\alpha^2 + 6\alpha - 4 = 4 \implies \alpha^2 - 6\alpha + 8 = 0$$

$$(\alpha - 2)(\alpha - 4) = 0 \implies \alpha = 4$$ (since $$\alpha > 2$$)

Computing the expression:

With $$\alpha = 4$$: $$B = \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix}$$, $$\vec{v} = \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix}$$

First compute $$B\vec{v}$$:

$$B\vec{v} = \begin{bmatrix} 4 - 24 + 16 \\ 4 - 16 + 12 \\ 16 - 32 + 16 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix}$$

Then $$\vec{v}^T B\vec{v} = [4, -8, 4] \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix} = -16$$

The answer is Option C: $$-16$$.

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): $$\boxed{4097}$$.

We have $$P^T = aP + (a-1)I$$ where $$P$$ is a $$3 \times 3$$ real matrix and $$a > 1$$.

Taking the transpose of both sides gives $$P = aP^T + (a-1)I$$.

Now substituting $$P^T = aP + (a-1)I$$ into this:

$$P = a[aP + (a-1)I] + (a-1)I = a^2P + a(a-1)I + (a-1)I = a^2P + (a-1)(a+1)I$$

$$P - a^2P = (a^2-1)I$$

$$P(1-a^2) = (a^2-1)I$$

$$P = -I$$

So $$|P| = |-I| = (-1)^3 = -1$$.

For a $$3 \times 3$$ matrix, we use the property $$|\text{Adj } P| = |P|^{n-1} = |P|^2 = (-1)^2 = 1$$.

Hence, the answer is $$|\text{Adj } P| = 1$$.

To find the values of $$\alpha$$and$$\beta$$, we can use the given equation for $$A^2$$to find an expression for$$A^4$$.

Given:

$$A^2 = 3A + \alpha I$$

$$A^4 = 21A + \beta I$$

Step 1: Express $$A^4$$in terms of $$A$$ and $$I$$ using the first equation.

Square both sides of the equation for $$A^2$$:

$$A^4 = (A^2)^2 = (3A + \alpha I)^2$$

$$A^4 = 9A^2 + 6\alpha A + \alpha^2 I$$

Step 2: Substitute $$A^2$$ back into the expanded equation.

$$A^4 = 9(3A + \alpha I) + 6\alpha A + \alpha^2 I$$

$$A^4 = 27A + 9\alpha I + 6\alpha A + \alpha^2 I$$

$$A^4 = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$$

Step 3: Compare this result with the given equation for $$A^4$$.

We are given that:

$$A^4 = 21A + \beta I$$

By equating the coefficients of $$A$$ and $$I$$ from both expressions, we get :

$$27 + 6\alpha = 21$$

$$9\alpha + \alpha^2 = \beta$$

Step 4: Solve for $$\alpha$$ and $$\beta$$.

From the first equation:

$$6\alpha = 21 - 27$$

$$\alpha = -1$$

Substitute $$\alpha = -1$$ into the second equation:

$$\beta = 9(-1) + (-1)^2$$

$$\beta = -9 + 1$$

$$\beta = -8$$

Final Answer

The values are $$\alpha = -1$$ and $$\beta = -8$$.

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 $$\alpha = \frac{1}{2} + 2 = \frac{5}{2}$$

To find the relation between the determinant d and the elements of the matrix A, we evaluate the given determinant condition:

$$\left| A - d(\text{Adj } A) \right| = 0$$

The matrix A and its determinant d are defined as:

$$A = \begin{pmatrix} m & n \\ p & q \end{pmatrix}, \quad d = |A| = mq - np$$

The adjoint of a 2x2 matrix is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:

$$\text{Adj } A = \begin{pmatrix} q & -n \\ -p & m \end{pmatrix}$$

---

Step 1: Express the matrix $$A - d(\text{Adj } A)$$

Multiplying the adjoint matrix by d:

$$d(\text{Adj } A) = \begin{pmatrix} dq & -dn \\ -dp & dm \end{pmatrix}$$

Now, subtracting this from matrix A:

$$A - d(\text{Adj } A) = \begin{pmatrix} m & n \\ p & q \end{pmatrix} - \begin{pmatrix} dq & -dn \\ -dp & dm \end{pmatrix} = \begin{pmatrix} m - dq & n(1 + d) \\ p(1 + d) & q - dm \end{pmatrix}$$

---

Step 2: Evaluate the determinant and set it to zero

$$\left| A - d(\text{Adj } A) \right| = \begin{vmatrix} m - dq & n(1 + d) \\ p(1 + d) & q - dm \end{vmatrix} = 0$$

Expanding the determinant:

$$(m - dq)(q - dm) - np(1 + d)^2 = 0$$

$$mq - m^2d - q^2d + mqd^2 - np(1 + d)^2 = 0$$

---

Step 3: Simplify using the definition of d

Grouping the terms with mq:

$$mq(1 + d^2) - d(m^2 + q^2) - np(1 + d)^2 = 0$$

We know that np = mq - d. Substituting this into the equation:

$$mq(1 + d^2) - d(m^2 + q^2) - (mq - d)(1 + 2d + d^2) = 0$$

$$mq + mqd^2 - dm^2 - dq^2 - (mq + 2mqd + mqd^2 - d - 2d^2 - d^3) = 0$$

Canceling out the common terms $$mq$$ and $$mqd^2$$:

$$-dm^2 - dq^2 - 2mqd + d + 2d^2 + d^3 = 0$$

Dividing the entire equation by $$-d$$ (since $$d \neq 0$$):

$$m^2 + q^2 + 2mq - 1 - 2d - d^2 = 0$$

---

Step 4: Form the perfect squares

Recognizing the algebraic expansions for $$(m + q)^2$$ and $$(1 + d)^2$$:

$$(m + q)^2 - (1 + 2d + d^2) = 0$$

$$(m + q)^2 - (1 + d)^2 = 0$$

$$(1 + d)^2 = (m + q)^2$$

Therefore, the final relationship is equal to $$(1 + d)^2 = (m + q)^2$$.

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.

1. Simplify the given determinant expression

The given condition is:

$$\left\vert{} \text{Adj}\left(2 \cdot \text{Adj}(2A^{-1})\right) \right\vert{} = 2^{84}$$

Using the property of the adjoint determinant $$\left\vert{} \text{Adj}(B) \right\vert{} = \vert{}B\vert{}^{n-1}$$ for a matrix of order $$n \times n$$, we can rewrite the equation by setting $$B = 2 \cdot \text{Adj}(2A^{-1})$$:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{}^{n-1} = 2^{84}$$

2. Evaluate the determinant inside the brackets

Using the scalar multiplication property $$\left\vert{} k \cdot B \right\vert{} = k^n \vert{}B\vert{}$$ for an $$n \times n$$ matrix, we expand the inner determinant:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{} = 2^n \left\vert{} \text{Adj}(2A^{-1}) \right\vert{}$$

Applying the adjoint property again to $$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{}$$ gives:

$$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{} = \left\vert{} 2A^{-1} \right\vert{}^{n-1}$$

3. Simplify the inverse matrix determinant

Using the properties $$\left\vert{} k \cdot A^{-1} \right\vert{} = k^n \left\vert{} A^{-1} \right\vert{}$$ and $$\left\vert{} A^{-1} \right\vert{} = \frac{1}{\vert{}A\vert{}}$$, we evaluate $$\left\vert{} 2A^{-1} \right\vert{}$$:

$$\left\vert{} 2A^{-1} \right\vert{} = 2^n \cdot \frac{1}{\vert{}A\vert{}}$$

Since we are given that $$\vert{}A\vert{} = 2$$, this simplifies to:

$$\left\vert{} 2A^{-1} \right\vert{} = 2^n \cdot \frac{1}{2} = 2^{n-1}$$

4. Substitute back to find the final equation

Now substitute $$\left\vert{} 2A^{-1} \right\vert{} = 2^{n-1}$$ back into the adjoint expression:

$$\left\vert{} \text{Adj}(2A^{-1}) \right\vert{} = (2^{n-1})^{n-1} = 2^{(n-1)^2}$$

Next, substitute this into the expression for the entire inner determinant:

$$\left\vert{} 2 \cdot \text{Adj}(2A^{-1}) \right\vert{} = 2^n \cdot 2^{(n-1)^2} = 2^{n + n^2 - 2n + 1} = 2^{n^2 - n + 1}$$

Finally, substitute this back into our original equation:

$$(2^{n^2 - n + 1})^{n-1} = 2^{84}$$

$$2^{(n^2 - n + 1)(n - 1)} = 2^{84}$$

5. Solve for n

Equating the exponents gives:

$$(n^2 - n + 1)(n - 1) = 84$$

Let us test integer values for n:

If n = 5:

$$(5^2 - 5 + 1)(5 - 1) = (25 - 5 + 1)(4) = 21 \cdot 4 = 84$$

Since this satisfies the equation, the value of n is 5.

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

$$A^n = I + nN + \binom{n}{2}N^2 = \begin{pmatrix} 1 & na & na + \binom{n}{2}ab \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{pmatrix}$$

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

$$48 + \frac{n(n-1)}{2} \cdot \frac{48}{n} \cdot \frac{96}{n} = 2160$$

which simplifies to

$$48 + \frac{(n-1) \cdot 48 \cdot 96}{2n} = 2160$$

$$\frac{2304(n-1)}{n} = 2112$$

$$2304n - 2304 = 2112n$$

$$192n = 2304 \implies n = 12$$

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.

Given,

$$A^5=B^5$$

and

$$A^3B^2=A^2B^3$$

Also,

$$A^2-B^2$$

is invertible.

From

$$A^3B^2=A^2B^3$$

we get

$$A^2B^2(A-B)=0$$

Now,

$$A^5-B^5=0$$

Factorizing,

$$(A-B)(A^4+A^3B+A^2B^2+AB^3+B^4)=0$$

Since powers of matrices commute here, write

$$A^4+A^3B+A^2B^2+AB^3+B^4$$

as

$$(A^2+B^2)(A^2+AB+B^2)-A^2B^2$$

Using

$$A^2B^2(A-B)=0,$$

we get

$$A^2B^2=0$$

Hence,

$$A^5-B^5=(A-B)(A^2+B^2)(A^2+AB+B^2)=0$$

But

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

is invertible.

Therefore,

$$A-B$$

is invertible.

Hence from

$$(A-B)(A^4+A^3B+A^2B^2+AB^3+B^4)=0$$

we get

$$A^4+A^3B+A^2B^2+AB^3+B^4=0$$

Now multiply by

$$A+B$$

Using algebraic identity,

$$A^5+B^5=0$$

But given

$$A^5=B^5$$

Therefore,

$$2A^5=0$$

$$A^5=0$$

Hence,

$$B^5=0$$

Thus both $$A$$ and $$B$$ are nilpotent matrices.

Therefore,

$$A^3+B^3$$

is also nilpotent.

Determinant of a nilpotent matrix is always zero.

Hence,

$$\boxed{0}$$

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 $$16$$.

Let $$M=\begin{pmatrix} 0&i\\ 1&0 \end{pmatrix}$$

We require

$$M^nA=A \qquad\forall A$$

This is possible only when $$M^n=I$$

Now,

$$M^2=\begin{pmatrix}0&i\\1&0\end{pmatrix}\begin{pmatrix}0&i\\1&0\end{pmatrix}=\begin{pmatrix}i&0\\0&i\end{pmatrix}=iI$$

Hence,

$$M^4=(iI)^2=-I$$ and $$M^8=(-I)^2=I$$

Therefore,

$$M^n=I \iff 8\mid n$$

Now count two-digit multiples of $$8.$$

Smallest two-digit multiple: $$16$$

Largest two-digit multiple: $$96$$

Thus number of terms is $$\frac{96-16}{8}+1$$

$$=10+1$$

$$=11$$

Hence, the required answer is $$\boxed{11}$$

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

$$ \sin\theta+\sin4\theta =2\cos\!\left(\dfrac{3\theta}{2}\right) =2\cos\!\left(\dfrac{3\pi}{10}\right). $$