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

The system: $$x+y+z=6$$, $$x+2y+5z=9$$, $$x+5y+\lambda z=\mu$$ has no solution when the determinant of coefficients is zero but the system is inconsistent.

Compute the determinant of the coefficient matrix: $$D = \begin{vmatrix}1&1&1\\1&2&5\\1&5&\lambda\end{vmatrix} = 1(2\lambda-25)-1(\lambda-5)+1(5-2) = 2\lambda-25-\lambda+5+3 = \lambda-17$$.

For no solution one requires $$D = 0 \implies \lambda = 17$$.

With $$\lambda = 17$$, the determinant of the augmented matrix replacing the first column by the constants becomes $$D_x = \begin{vmatrix}6&1&1\\9&2&5\\\mu&5&17\end{vmatrix} = 6(34-25)-1(153-5\mu)+1(45-2\mu) = 54-153+5\mu+45-2\mu = 3\mu-54$$.

Inconsistency requires $$D_x \neq 0 \implies 3\mu \neq 54 \implies \mu \neq 18$$.

Thus the system has no solution when $$\lambda = 17$$ and $$\mu \neq 18$$.

The correct answer is Option 3: $$\lambda = 17, \mu \neq 18$$.

The determinant is

$$y(x)= \begin{vmatrix} \sin x & \cos x & \sin x+\cos x+1\\ 27 & 28 & 27\\ 1 & 1 & 1 \end{vmatrix}$$

We simplify the determinant by a column operation. Replace the third column by $$C_3 \;-\; C_1 \;-\; C_2$$ (this does not change the value of a determinant):

$$ C_3 \longrightarrow C_3-C_1-C_2 \quad\Longrightarrow\quad y(x)= \begin{vmatrix} \sin x & \cos x & 1\\ 27 & 28 & -28\\ 1 & 1 & -1 \end{vmatrix} $$

Now expand along the first row:

$$ y(x)=\sin x\, \bigl(28\cdot(-1)-(-28)\cdot1\bigr) -\cos x\, \bigl(27\cdot(-1)-(-28)\cdot1\bigr) +1\, \bigl(27\cdot1-28\cdot1\bigr) $$

Simplifying each bracket:

$$ \begin{aligned} 28(-1)-(-28)(1) &= -28+28=0,\\ 27(-1)-(-28)(1) &= -27+28=1,\\ 27(1)-28(1) &= -1. \end{aligned} $$

Hence

$$y(x)=\sin x\cdot0-\cos x\cdot1+1\cdot(-1)=-\cos x-1.$$

Differentiate twice with respect to $$x$$.

First derivative: $$y'(x)=\frac{d}{dx}\bigl(-\cos x-1\bigr)=\sin x.$$

Second derivative: $$y''(x)=\frac{d}{dx}\bigl(\sin x\bigr)=\cos x.$$

Now compute $$y''+y$$:

$$y''(x)+y(x)=\cos x+\bigl(-\cos x-1\bigr)=-1.$$

Thus $$\displaystyle\frac{d^{2}y}{dx^{2}}+y=-1.$$

The expression is the constant $$-1$$ for all $$x\in\mathbb{R}$$. So the correct choice is Option A.

 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.

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ($$D$$) and the determinants associated with each variable ($$D_x, D_y, D_z$$) must all equal zero.

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Step 1: Compute the main determinant $$D$$ and set it to zero

$$D = \begin{vmatrix} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$D = 3(\alpha(1) - (-1)(2)) - 1(2(1) - (-1)(1)) + \beta(2(2) - \alpha(1)) = 0$$

$$D = 3(\alpha + 2) - 1(3) + \beta(4 - \alpha) = 0$$

$$3\alpha + 6 - 3 + 4\beta - \alpha\beta = 0 \implies 3\alpha + 4\beta - \alpha\beta + 3 = 0$$

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Step 2: Compute the determinant $$D_y$$ and set it to zero

To simplify calculations, we substitute the constant column into the second column to find $$D_y$$:

$$D_y = \begin{vmatrix} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$D_y = 3(-3(1) - (-1)(4)) - 3(2(1) - (-1)(1)) + \beta(2(4) - (-3)(1)) = 0$$

$$D_y = 3(-3 + 4) - 3(2 + 1) + \beta(8 + 3) = 0$$

$$3(1) - 3(3) + 11\beta = 0 \implies 3 - 9 + 11\beta = 0$$

$$11\beta - 6 = 0 \implies \beta = \frac{6}{11}$$

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Step 3: Solve for $$\alpha$$

Substituting the value of $$\beta = \frac{6}{11}$$ back into the equation for $$D = 0$$:

$$3\alpha + 4\left(\frac{6}{11}\right) - \alpha\left(\frac{6}{11}\right) + 3 = 0$$

$$3\alpha - \frac{6}{11}\alpha + \frac{24}{11} + 3 = 0$$

$$\frac{27}{11}\alpha + \frac{57}{11} = 0 \implies 27\alpha = -57 \implies \alpha = -\frac{57}{27} = -\frac{19}{9}$$

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Step 4: Evaluate the required expression

Now, substitute the values of $$\alpha = -\frac{19}{9}$$ and $$\beta = \frac{6}{11}$$ into the given expression:

$$22\beta - 9\alpha = 22\left(\frac{6}{11}\right) - 9\left(-\frac{19}{9}\right)$$

$$22\beta - 9\alpha = 2(6) + 19 = 12 + 19 = 31$$

Therefore, the value of $$22\beta - 9\alpha$$ is equal to 31.

The given function is:

$$f(x) = \begin{vmatrix} 1+\sin^{2}x & \cos^{2}x & 4\sin 4x \\ \sin^{2}x & 1+\cos^{2}x & 4\sin 4x \\ \sin^{2}x & \cos^{2}x & 1+4\sin 4x \end{vmatrix}$$

To simplify the determinant, perform row operations. Subtract row 3 from row 1 and row 2:

$$R_1 - R_3 = \begin{bmatrix} (1+\sin^{2}x) - \sin^{2}x & \cos^{2}x - \cos^{2}x & 4\sin 4x - (1+4\sin 4x) \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \end{bmatrix}$$

$$R_2 - R_3 = \begin{bmatrix} \sin^{2}x - \sin^{2}x & (1+\cos^{2}x) - \cos^{2}x & 4\sin 4x - (1+4\sin 4x) \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \end{bmatrix}$$

The determinant remains unchanged, so:

$$f(x) = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \sin^{2}x & \cos^{2}x & 1+4\sin 4x \end{vmatrix}$$

Expand along the first row:

$$f(x) = 1 \cdot \begin{vmatrix} 1 & -1 \\ \cos^{2}x & 1+4\sin 4x \end{vmatrix} - 0 \cdot \begin{vmatrix} 0 & -1 \\ \sin^{2}x & 1+4\sin 4x \end{vmatrix} + (-1) \cdot \begin{vmatrix} 0 & 1 \\ \sin^{2}x & \cos^{2}x \end{vmatrix}$$

The second term is zero. Compute the other determinants:

First determinant: $$\begin{vmatrix} 1 & -1 \\ \cos^{2}x & 1+4\sin 4x \end{vmatrix} = 1 \cdot (1+4\sin 4x) - (-1) \cdot \cos^{2}x = 1 + 4\sin 4x + \cos^{2}x$$

Second determinant: $$\begin{vmatrix} 0 & 1 \\ \sin^{2}x & \cos^{2}x \end{vmatrix} = 0 \cdot \cos^{2}x - 1 \cdot \sin^{2}x = -\sin^{2}x$$

So:

$$f(x) = (1 + 4\sin 4x + \cos^{2}x) - (-\sin^{2}x) = 1 + 4\sin 4x + \cos^{2}x + \sin^{2}x$$

Using the identity $$\sin^{2}x + \cos^{2}x = 1$$:

$$f(x) = 1 + 4\sin 4x + 1 = 2 + 4\sin 4x$$

The range of $$\sin 4x$$ is $$[-1, 1]$$, so:

Maximum value of $$f(x)$$ occurs when $$\sin 4x = 1$$:

$$M = 2 + 4(1) = 6$$

Minimum value of $$f(x)$$ occurs when $$\sin 4x = -1$$:

$$m = 2 + 4(-1) = -2$$

Now compute $$M^4 - m^4$$:

$$M^4 = 6^4 = 1296$$

$$m^4 = (-2)^4 = 16$$

$$M^4 - m^4 = 1296 - 16 = 1280$$

Alternatively, using the difference of squares:

$$M^4 - m^4 = (M^2)^2 - (m^2)^2 = (M^2 - m^2)(M^2 + m^2)$$

$$M^2 = 36, \quad m^2 = 4$$

$$M^2 - m^2 = 36 - 4 = 32$$

$$M^2 + m^2 = 36 + 4 = 40$$

$$M^4 - m^4 = 32 \times 40 = 1280$$

The value of $$M^4 - m^4$$ is 1280.

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero ($$\Delta = 0$$) and the system must be consistent ($$\Delta_z = 0$$). The coefficient matrix $$A$$ is

$$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{bmatrix}$$

so its determinant is computed by expanding along the first row:

$$\Delta = 1 \cdot \det \begin{bmatrix} 3 & a \\ -3 & b \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} + 2 \cdot \det \begin{bmatrix} 2 & 3 \\ -1 & -3 \end{bmatrix}$$

Computing each $$2 \times 2$$ determinant gives:

$$\Delta = 1(3b + 3a) - 1(2b + a) + 2(-6 + 3) = 2a + b - 6$$

Setting $$\Delta = 0$$ gives $$2a + b = 6$$. Next, computing $$\Delta_z$$ by replacing the third column with the constant terms:

$$\Delta_z = \det \begin{bmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{bmatrix}$$

Expanding along the first row yields:

$$\Delta_z = 1 \cdot (6b + 3a + 3) - 1 \cdot (4b + a + 1) + 6 \cdot (-3) = 2a + 2b - 16$$

Setting $$\Delta_z = 0$$ gives $$2a + 2b = 16$$, which simplifies to $$a + b = 8$$.

We now have a system of two equations, $$2a + b = 6$$ and $$a + b = 8$$. Subtracting the second equation from the first gives $$a = -2$$. Substituting this value into the second equation yields $$b = 10$$.

Finally, substituting these values into the required expression:

$$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16$$

Therefore, the final answer is $$16$$.

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.

Given that $$A$$ is a $$3 \times 3$$ matrix. For an $$n \times n$$ matrix, $$|\text{adj}(A)| = |A|^{n-1}$$.

For $$n = 3$$: $$|\text{adj}(A)| = |A|^2$$, $$|\text{adj}(\text{adj}(A))| = |A|^{(n-1)^2} = |A|^4$$, and $$|\text{adj}(\text{adj}(\text{adj}(A)))| = |A|^{(n-1)^3} = |A|^8$$.

Given $$|A|^8 = 81 = 3^4$$, so $$|A|^2 = 81^{1/4} = 3$$.

Now for the set $$S$$, we need: $$|\text{adj}(\text{adj}(A))|^{\frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$$

$$|A|^{4 \cdot \frac{(n-1)^2}{2}} = |A|^{3n^2 - 5n - 4}$$

$$|A|^{2(n-1)^2} = |A|^{3n^2 - 5n - 4}$$

Since $$|A| \neq 0$$ and $$|A| \neq \pm 1$$, equating exponents:

$$2(n^2 - 2n + 1) = 3n^2 - 5n - 4$$

$$2n^2 - 4n + 2 = 3n^2 - 5n - 4$$

$$n^2 - n - 6 = 0$$, giving $$(n-3)(n+2) = 0$$

So $$S = \{3, -2\}$$.

$$\sum_{n \in S} |A^{n^2+n}| = |A|^{9+3} + |A|^{4-2} = |A|^{12} + |A|^{2}$$

Since $$|A|^2 = 3$$: $$|A|^{12} = (|A|^2)^6 = 3^6 = 729$$ and $$|A|^2 = 3$$.

Sum $$= 729 + 3 = 732$$.

Hence, the correct answer is Option D.

Let $$t=\frac{\sin x}{x}\,,$$ so that $$\displaystyle\lim_{x\to 0}t=1$$ because $$\frac{\sin x}{x}\to 1$$ as $$x\to 0$$.

With this notation the determinant becomes
$$f(x)=\left|\begin{matrix}a+t & 1 & b\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Perform the row operation $$R_1\rightarrow R_1-R_2$$ (subtract the second row from the first):
$$f(x)=\left|\begin{matrix}t & -t & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Factor out $$t$$ from the first row:
$$f(x)=t\;\left|\begin{matrix}1 & -1 & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Call the remaining determinant $$D(t)$$, i.e.
$$D(t)=\left|\begin{matrix}1 & -1 & 0\\ a & 1+t & b\\ a & 1 & b+t\end{matrix}\right|.$$

Expand $$D(t)$$ along the first row:
$$\begin{aligned} D(t)=&\;1\;\left|\begin{matrix}1+t & b\\ 1 & b+t\end{matrix}\right| \;-\;(-1)\;\left|\begin{matrix}a & b\\ a & b+t\end{matrix}\right| \;+\;0\\[4pt] =&\;(1+t)(b+t)-b + a\,t. \end{aligned}$$

Simplify the expression:
$$(1+t)(b+t)-b = (1+t)(b+t)-b = (1+t)(b+t)-b.$$
Therefore
$$D(t)=(1+t)(b+t)-b+a\,t.$$

Now take the limit as $$x\to 0$$, i.e. $$t\to 1$$:
$$\begin{aligned} \lim_{x\to 0}f(x) &=\lim_{t\to 1}\;t\,D(t)\\ &=1\;\bigl[(1+1)(b+1)-b+a\cdot 1\bigr]\\ &=(2)(b+1)-b+a\\ &=2b+2-b+a\\ &=a+b+2. \end{aligned}$$

Hence
$$\lambda+\mu a+\nu b=a+b+2,$$
so $$\lambda=2,\;\mu=1,\;\nu=1.$$

The required value is
$$(\lambda+\mu+\nu)^2=(2+1+1)^2=4^2=16.$$

Therefore, the correct option is Option A (16).

We must evaluate $$\det\bigl(\text{adj}\bigl(\text{adj}(3A)\bigr)\bigr)$$, where the given matrix is
$$A=\begin{bmatrix}2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q\end{bmatrix}.$$

Case 1 : Relating adjugates to determinants

For any non-singular $$n\times n$$ matrix $$M$$, two standard identities are

$$\det\bigl(\text{adj}(M)\bigr)=(\det M)^{\,n-1}$$

$$\text{adj}\bigl(\text{adj}(M)\bigr)=(\det M)^{\,n-2}\,M$$  for $$n\gt1$$.

Here $$n=3$$ and $$M=3A$$. Therefore

$$\text{adj}\bigl(\text{adj}(3A)\bigr)=\bigl(\det(3A)\bigr)^{\,3-2}\,(3A)=\det(3A)\,(3A).$$

Case 2 : Determinant of the obtained matrix

Let $$B=\det(3A)\,(3A)$$. Since multiplying a $$3\times3$$ matrix by a scalar $$c$$ multiplies its determinant by $$c^{\,3}$$, we get

$$\det(B)=\bigl(\det(3A)\bigr)^{3}\,\det(3A)=\bigl(\det(3A)\bigr)^{4}.$$

Case 3 : Evaluating $$\det(3A)$$

First compute $$\det(A)$$. Apply the row operations $$R_2\leftarrow R_2-2R_1$$ and $$R_3\leftarrow R_3-3R_1$$:

$$\begin{bmatrix} 2 & 2+p & 2+p+q\\ 4 & 6+2p & 8+3p+2q\\ 6 & 12+3p & 20+6p+3q \end{bmatrix} \;\longrightarrow\; \begin{bmatrix} 2 & 2+p & 2+p+q\\ 0 & 2 & 4+p\\ 0 & 6 & 14+3p \end{bmatrix}.$$ Only elementary row replacements were used, so the determinant is unchanged. Expanding along the first column,

$$\det(A)=2\begin{vmatrix}2 & 4+p\\ 6 & 14+3p\end{vmatrix} =2\bigl[(2)(14+3p)-(4+p)(6)\bigr]$$

$$=2\bigl[(28+6p)-(24+6p)\bigr]=2\,(4)=8.$$

Hence $$\det(A)=8=2^{3}.$$

Now, $$\det(3A)=3^{3}\det(A)=27\times8=216=2^{3}\,3^{3}.$$

Case 4 : Final value of the required determinant

Substituting $$\det(3A)=2^{3}3^{3}$$ into Case 2 gives

$$\det\bigl(\text{adj}(\text{adj}(3A))\bigr)=\bigl(2^{3}3^{3}\bigr)^{4}=2^{12}\,3^{12}.$$

Thus $$m=12,\;n=12$$ and

$$m+n=24.$$

The correct choice is Option B (24).

We are given

$$A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}, \qquad \det(A) = -4.$$

Subtracting the identity matrix $$I$$, we find

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

Step 1: Determine $$a$$ using $$\det(A) = -4$$.

Compute the determinant:

$$\det(A) = 0 \;-\; a\begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix} \;+\; 1\begin{vmatrix} 2 & 0 \\ a & 1 \end{vmatrix}$$

$$\det(A) = -a(2\cdot1 - 0\cdot a) + (2\cdot1 - 0\cdot a) = -2a + 2 = 2(1-a).$$

Given $$\det(A) = -4$$, we have

$$2(1-a) = -4 \;\Longrightarrow\; 1-a = -2 \;\Longrightarrow\; a = 3.$$

Step 2: Set up the required determinant.

We need

$$D = \det\!\bigl((a+1)\,\operatorname{adj}((a-1)A)\bigr).$$

Property used: For an $$n \times n$$ matrix $$B$$,

$$\det\!\bigl(\operatorname{adj}(B)\bigr)=\det(B)^{\,n-1}.$$

Here $$n = 3,$$ so $$\det(\operatorname{adj}(B)) = \det(B)^{2}.$$

Also, for a scalar $$k,$$

$$\det(kB) = k^{\,n}\det(B).$$

Step 3: Apply the properties.

Let $$B = (a-1)A.$$ Then

$$D = (a+1)^{3}\,\det\!\bigl(\operatorname{adj}(B)\bigr) = (a+1)^{3}\,\det(B)^{2}.$$

Compute $$\det(B):$$

$$\det(B) = \det\bigl((a-1)A\bigr) = (a-1)^{3}\det(A).$$

Step 4: Insert $$a = 3$$ and $$\det(A) = -4.$

$$a+1 = 4,\quad a-1 = 2.$$

$$\det(B) = 2^{3}(-4) = 8(-4) = -32.$$

Hence

$$D = 4^{3}\,(-32)^{2} = 64 $$\times$$ 1024 = 65536.$$

Step 5: Express $$D$$ as $$2^{m}3^{n}.$$

$$65536 = 2^{16},$$ so $$m = 16,\; n = 0.$$

Step 6: Compute $$m+n.$$

$$m+n = 16 + 0 = 16.$$

Therefore, the required value is $$16$$, which corresponds to Option D.

The given system is

$$\begin{aligned} x+5y-z &= 1 \quad\; -(1)\\ 4x+3y-3z &= 7 \quad -(2)\\ 24x+y+\lambda z &= \mu \quad -(3) \end{aligned}$$

For infinitely many solutions the three equations must be dependent, i.e. equation $$(3)$$ must be a linear combination of $$(1)$$ and $$(2)$$ with the same combination on the right-hand sides.

Let $$\alpha,\,\beta$$ be real numbers such that

$$\alpha(x+5y-z)+\beta(4x+3y-3z)=24x+y+\lambda z \quad -(4)$$

Matching the coefficients of $$x,\;y,\;z$$ in $$(4)$$ gives

$$\begin{aligned} \alpha+4\beta &= 24 \quad -(5)\\ 5\alpha+3\beta &= 1 \quad -(6)\\ -\alpha-3\beta &= \lambda \quad -(7) \end{aligned}$$

Solving $$(5)$$ and $$(6)$$:
Multiply $$(5)$$ by $$5$$ ⇒ $$5\alpha+20\beta=120$$.
Subtract $$(6)$$ ⇒ $$17\beta = 119 \;\Rightarrow\; \beta = 7$$.
From $$(5)$$ ⇒ $$\alpha = 24-4\beta = 24-28 = -4$$.

Then $$(7)$$ gives $$\lambda = (-\alpha-3\beta)=4-21=-17$$.

The right-hand side of $$(3)$$ must also match:
$$\mu = \alpha\cdot1 + \beta\cdot7 = (-4)\cdot1 + 7\cdot7 = -4+49 = 45$$.

Hence infinitely many solutions occur only for
$$\boxed{\lambda=-17,\;\mu = 45}$$.

With these values, equation $$(3)$$ is redundant and the system reduces to $$(1)$$(2). Solve it parametrically.

From $$(1)$$: $$x = 1 - 5y + z \quad -(8)$$.

Substitute $$(8)$$ in $$(2)$$:

$$4(1-5y+z) + 3y - 3z = 7$$ $$4 - 20y + 4z + 3y - 3z = 7$$ $$4 - 17y + z = 7$$ $$\Rightarrow\; z = 3 + 17y \quad -(9)$$.

Insert $$(9)$$ into $$(8)$$:

$$x = 1 - 5y + (3+17y) = 4 + 12y \quad -(10)$$.

Let $$y = t$$ (any integer). Then

$$x = 4 + 12t,\; y = t,\; z = 3 + 17t \quad -(11)$$

The required condition is $$7 \le x+y+z \le 77$$.

Using $$(11)$$:
$$x+y+z = (4+12t)+t+(3+17t) = 7 + 30t$$.

So

$$7 \le 7 + 30t \le 77 \;\Longrightarrow\; 0 \le 30t \le 70$$
$$\Rightarrow\; 0 \le t \le \frac{70}{30} = \frac{7}{3}$$.

Since $$t$$ is an integer, $$t = 0,1,2$$. Thus there are

$$\boxed{3}$$

integer triples $$(x,y,z)$$ satisfying all the given conditions.

Hence the correct option is Option A (3).

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 $$\boxed{2}$$.

Let the matrix be $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$ where each entry comes from the set $$S=\{2,3,6,9\}$$.

A $$2\times2$$ matrix is singular when its determinant is zero, i.e. when
$$\det\begin{pmatrix}a & b \\ c & d\end{pmatrix}=ad-bc=0 \; \Longleftrightarrow \; ad=bc.$$ Thus we must count the ordered quadruples $$(a,b,c,d)\in S^{4}$$ satisfying $$ad=bc.$$

Step 1 - List every possible ordered pair from $$S$$ together with its product $$p=xy$$.

Because order matters, $$(x,y)$$ and $$(y,x)$$ are counted separately.

$$\begin{array}{c|c} (x,y) & p=xy \\ \hline (2,2) & 4 \\ (2,3),\,(3,2) & 6 \\ (3,3) & 9 \\ (2,6),\,(6,2) & 12 \\ (2,9),\,(3,6),\,(6,3),\,(9,2) & 18 \\ (3,9),\,(9,3) & 27 \\ (6,6) & 36 \\ (6,9),\,(9,6) & 54 \\ (9,9) & 81 \end{array}$$

Denote by $$n_p$$ the number of ordered pairs $$(x,y)$$ whose product equals $$p$$.
From the table we obtain

$$n_4=1,\; n_6=2,\; n_9=1,\; n_{12}=2,\; n_{18}=4,\; n_{27}=2,\; n_{36}=1,\; n_{54}=2,\; n_{81}=1.$$

Step 2 - Form the matrix.
For a fixed product $$p$$, we can pick $$(a,d)$$ in $$n_p$$ ways and independently pick $$(b,c)$$ in another $$n_p$$ ways, giving $$n_p^2$$ matrices with determinant zero corresponding to that product.

Step 3 - Add over all possible products:

$$\text{Number of singular matrices}= \sum_{p} n_p^{2}$$ $$=1^{2}+2^{2}+1^{2}+2^{2}+4^{2}+2^{2}+1^{2}+2^{2}+1^{2}$$ $$=1+4+1+4+16+4+1+4+1$$ $$=36.$$

Hence, the total number of singular $$2\times2$$ matrices whose elements are chosen from $$\{2,3,6,9\}$$ is $$36$$.

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.

For infinitely many solutions, the third equation must be a linear combination of the first two.

Let (iii) = $$\alpha$$(i) + $$\beta$$(ii). Matching coefficients:

$$2\alpha + 5\beta = 100$$ ... (A), $$-\alpha + \lambda\beta = -47$$ ... (B), $$\alpha + 3\beta = \mu$$ ... (C), $$4\alpha + 12\beta = 212$$ ... (D)

From (D): $$\alpha + 3\beta = 53$$, so $$\mu = 53$$.

From (A) and $$\alpha = 53 - 3\beta$$: $$106 - 6\beta + 5\beta = 100$$, giving $$\beta = 6$$ and $$\alpha = 35$$.

From (B): $$-35 + 6\lambda = -47$$, so $$\lambda = -2$$.

$$\mu - 2\lambda = 53 - 2(-2) = 57$$.

The correct answer is Option 1: 57.

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.

For a system of three linear equations to possess infinitely many solutions,
the coefficient matrix $$A$$ must be singular (determinant $$0$$) and the augmented matrix must have the same rank as $$A$$ (that rank will then be $$2$$).

Write the coefficient matrix and its determinant:

$$A=\begin{vmatrix}2 & \lambda & 3\\ 3 & 2 & -1\\ 4 & 5 & \mu\end{vmatrix}$$

Expanding along the first row,

$$\begin{aligned} \det A &= 2\begin{vmatrix}2 & -1\\ 5 & \mu\end{vmatrix} -\lambda\begin{vmatrix}3 & -1\\ 4 & \mu\end{vmatrix} +3\begin{vmatrix}3 & 2\\ 4 & 5\end{vmatrix}\\[4pt] &= 2(2\mu+5)-\lambda(3\mu+4)+3(15-8)\\[4pt] &= 4\mu+10-\lambda(3\mu+4)+21\\[4pt] &= 4\mu+31-\lambda(3\mu+4). \end{aligned}$$

Setting $$\det A = 0$$ gives

$$\lambda(3\mu+4)=4\mu+31 \quad -(1)$$

Because $$\det A = 0$$, the three rows are linearly dependent.
Assume the third row is a linear combination of the first two:

$$R_3 = pR_1 + qR_2$$

This yields four scalar equations (for the coefficients of $$x,\,y,\,z$$ and the constants):

$$\begin{aligned} 2p+3q &= 4 \quad -(2)\\ \lambda p + 2q &= 5 \quad -(3)\\ 3p - q &= \mu \quad -(4)\\ 5p + 7q &= 9 \quad -(5) \end{aligned}$$

Solve equations $$(2)$$ and $$(3)$$ for $$p,\,q$$.
From $$(2):$$ $$p=\dfrac{4-3q}{2}.$$
Insert this into $$(3):$$ $$5 = \lambda\left(\dfrac{4-3q}{2}\right)+2q \;\;\Longrightarrow\;\; (-3\lambda+4)q = 10-4\lambda.$$

Set $$D = 4-3\lambda$$ (note $$D\neq0$$ for rank $$2$$). Then

$$q=\dfrac{10-4\lambda}{D},\qquad p=\dfrac{-7}{D}.$$

Check the constant equation $$(5)$$ for consistency:

$$5p+7q \;=\; 5\!\left(\dfrac{-7}{D}\right)+7\!\left(\dfrac{10-4\lambda}{D}\right) = \dfrac{35-28\lambda}{D}.$$

For infinite solutions this must equal $$9$$:

$$\dfrac{35-28\lambda}{D}=9 \;\;\Longrightarrow\;\; 35-28\lambda = 9(4-3\lambda) = 36-27\lambda.$$

Simplifying gives $$\lambda = -1.$$

Now substitute $$\lambda=-1$$ into $$(1)$$ to find $$\mu$$:

$$(-1)(3\mu+4)=4\mu+31 \;\;\Longrightarrow\;\; -3\mu-4 = 4\mu+31 \;\;\Longrightarrow\;\; 7\mu = -35 \;\;\Longrightarrow\;\; \mu = -5.$$

Finally,

$$(\lambda^2+\mu^2)=(-1)^2+(-5)^2 = 1+25 = 26.$$

Hence the required value is $$26$$, corresponding to Option C.

Matrix $$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$$ where each $$a_{ij} \in \{0, 1\}$$.

Total matrices = $$2^4 = 16$$.

$$\det(A) = a_{11}a_{22} - a_{12}a_{21}$$.

Possible values of det(A): -1, 0, 1.

Let us count each case:

det = 1: $$a_{11}a_{22} = 1$$ and $$a_{12}a_{21} = 0$$. So $$a_{11}=a_{22}=1$$ and at least one of $$a_{12}, a_{21}$$ is 0. Number of ways = 1 × 3 = 3. But we need $$a_{11}a_{22} - a_{12}a_{21} = 1$$, which means $$a_{11}a_{22}=1, a_{12}a_{21}=0$$: 3 cases.

det = -1: $$a_{11}a_{22} = 0$$ and $$a_{12}a_{21} = 1$$. So $$a_{12}=a_{21}=1$$ and at least one of $$a_{11}, a_{22}$$ is 0. Number of ways = 3 × 1 = 3.

det = 0: Remaining cases = 16 - 3 - 3 = 10.

Now computing variance:

$$E(X) = \frac{1}{16}[3(1) + 10(0) + 3(-1)] = 0$$

$$E(X^2) = \frac{1}{16}[3(1) + 10(0) + 3(1)] = \frac{6}{16} = \frac{3}{8}$$

$$Var(X) = E(X^2) - [E(X)]^2 = \frac{3}{8} - 0 = \frac{3}{8}$$

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

The matrix is $$A=\begin{pmatrix}0&1&c\\1&a&d\\1&b&e\end{pmatrix}$$ with $$a,b,c,d,e\in\{0,1\}$$.

First find an explicit expression for the determinant.

Expanding along the first row:
$$|A| \;=\;0\,(a e-d b)\;-\;1\,(1\cdot e-d\cdot 1)\;+\;c\,(1\cdot b-a\cdot 1)$$
$$\Longrightarrow\;|A| \;=\;-(e-d)+c(b-a).$$

Define two simpler variables:
$$x=d-e,\qquad y=b-a,$$
so that
$$|A| \;=\;x+c\,y.$$ Because each entry is either 0 or 1, $$x,\;y\in\{-1,0,1\}.$$ Specifically

• $$x=1$$ when $$(d,e)=(1,0),$$
• $$x=0$$ when $$(d,e)=(0,0)\ \text{or}\ (1,1),$$
• $$x=-1$$ when $$(d,e)=(0,1).$$
• $$y=1$$ when $$(b,a)=(1,0),$$
• $$y=0$$ when $$(b,a)=(0,0)\ \text{or}\ (1,1),$$
• $$y=-1$$ when $$(b,a)=(0,1).$$

The requirement is $$|A|=x+c\,y=\pm1.$$ Treat the two cases for $$c.$$

Case 1: $$c=0$$

Then $$|A|=x.$$ For $$|A|=\pm1$$ we need $$x=\pm1.$$

• $$x=1: (d,e)=(1,0)$$
• $$x=-1: (d,e)=(0,1)$$

That gives 2 choices for $$(d,e).$$

The pair $$(a,b)$$ is unrestricted (4 possibilities).

Total in this case: $$1\;(c)\times2\;(d,e)\times4\;(a,b)=8.$$

Case 2: $$c=1$$

Now $$|A|=x+y.$$ We need $$x+y=\pm1.$$ The admissible ordered pairs $$(x,y)$$ and their counts are

• $$(-1,0)$$: 1 way for $$x$$ × 2 ways for $$y$$ = 2
• $$(0,1)$$: 2 ways for $$x$$ × 1 way for $$y$$ = 2
• $$(0,-1)$$: 2 ways for $$x$$ × 1 way for $$y$$ = 2
• $$(1,0)$$: 1 way for $$x$$ × 2 ways for $$y$$ = 2

Adding these, we have 8 matrices when $$c=1.$$

Hence total in this case: $$1\;(c)\times8=8.$$

Adding the two cases:
$$8+8=16.$$

Therefore, the set $$S$$ contains 16 matrices.

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.

System: $$x + y + z = 5$$, $$x + 2y + \lambda^2 z = 9$$, $$x + 3y + \lambda z = \mu$$.

Determinant: $$D = \begin{vmatrix} 1&1&1\\1&2&\lambda^2\\1&3&\lambda\end{vmatrix} = 1(2\lambda-3\lambda^2) - 1(\lambda-\lambda^2) + 1(3-2) = 2\lambda-3\lambda^2-\lambda+\lambda^2+1 = -2\lambda^2+\lambda+1 = -(2\lambda^2-\lambda-1) = -(2\lambda+1)(\lambda-1)$$

$$D = 0$$ when $$\lambda = 1$$ or $$\lambda = -1/2$$.

Option (3): "System has unique solution if $$\lambda \neq 1$$ and $$\mu \neq 13$$." But $$D = 0$$ also when $$\lambda = -1/2$$, not just $$\lambda = 1$$. So for unique solution we need $$\lambda \neq 1$$ AND $$\lambda \neq -1/2$$. The statement ignores $$\lambda = -1/2$$ case.

This statement is NOT correct — if $$\lambda = -1/2$$ and $$\mu \neq 13$$, $$D = 0$$ so no unique solution.

The answer is Option (3): The statement that is NOT correct.

Expanding the given determinant or performing row operations ($$R_2 \to R_2-R_1, R_3 \to R_3-R_1$$) and dividing the resulting equation by $$(\alpha-a)(\beta-b)(\gamma-c)$$ yields:

$$\frac{\alpha}{\alpha-a} + \frac{b}{\beta-b} + \frac{c}{\gamma-c} = 0$$

Rewrite the first term as $$1 + \frac{a}{\alpha-a}$$:

$$\frac{a}{\alpha-a} + \frac{b}{\beta-b} + \frac{c}{\gamma-c} = -1$$

The required expression uses $$\gamma$$ instead of $$c$$ in the final numerator. Rewrite $$\frac{\gamma}{\gamma-c}$$ as $$1 + \frac{c}{\gamma-c}$$:

$$\frac{a}{\alpha-a} + \frac{b}{\beta-b} + \left(1 + \frac{c}{\gamma-c}\right) = 1 + (-1) = 0$$

$$f(0) = \begin{vmatrix}0&1&1\\2&0&6\\0&4&-2\end{vmatrix} = 0(0-24)-1(-4-0)+1(8-0) = 4+8 = 12$$.

For $$f'(0)$$, differentiate the determinant by differentiating each row separately. At $$x=0$$:

Row derivatives: $$(0, 0, 3)$$, $$(0, 2, 0)$$, $$(-1, 0, 0)$$.

$$f'(0) = \begin{vmatrix}0&0&3\\2&0&6\\0&4&-2\end{vmatrix} + \begin{vmatrix}0&1&1\\0&2&0\\0&4&-2\end{vmatrix} + \begin{vmatrix}0&1&1\\2&0&6\\-1&0&0\end{vmatrix}$$

First: $$0 - 0 + 3(8) = 24$$. Second: $$0 - 1(0) + 1(0) = 0$$. Third: $$0 - 1(0+6) + 1(0+0) = -6$$.

$$f'(0) = 24 + 0 - 6 = 18$$.

$$2f(0)+f'(0) = 24+18 = 42$$.

Given matrix $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$$ and the condition $$|2A|^3 = 2^{21}$$, with $$\alpha, \beta \in \mathbb{Z}$$.

First, recall that for an $$n \times n$$ matrix, $$|kA| = k^n |A|$$. Here, $$A$$ is a 3x3 matrix, so $$n=3$$. Thus, $$|2A| = 2^3 |A| = 8 |A|$$.

Substitute into the given condition: $$(8 |A|)^3 = 2^{21}$$. Since $$8 = 2^3$$, this becomes: $$(2^3 |A|)^3 = 2^{21}$$. Simplifying the exponents: $$2^{9} |A|^3 = 2^{21}$$.

Divide both sides by $$2^9$$: $$|A|^3 = 2^{12}$$. Taking the cube root: $$|A| = 2^{4} = 16$$. Therefore, the determinant of $$A$$ is 16.

Now, compute the determinant of $$A$$. Expanding along the first row: $$|A| = 1 \cdot \begin{vmatrix} \alpha & \beta \\ \beta & \alpha \end{vmatrix} - 0 \cdot (\ldots) + 0 \cdot (\ldots) = \alpha^2 - \beta^2$$.

So, $$\alpha^2 - \beta^2 = 16$$. Factorizing: $$(\alpha - \beta)(\alpha + \beta) = 16$$. Since $$\alpha$$ and $$\beta$$ are integers, both $$\alpha - \beta$$ and $$\alpha + \beta$$ are integers and must be even (as their product is even and both must have the same parity).

Set $$x = \alpha - \beta$$ and $$y = \alpha + \beta$$, so $$x \cdot y = 16$$, and both $$x$$ and $$y$$ are even integers. The even factor pairs of 16 are: $$(2,8)$$, $$(4,4)$$, $$(8,2)$$, $$(-2,-8)$$, $$(-4,-4)$$, $$(-8,-2)$$.

Solve for $$\alpha$$ and $$\beta$$ in each pair using $$\alpha = \frac{x + y}{2}$$ and $$\beta = \frac{y - x}{2}$$:

• For $$(2,8)$$: $$\alpha = \frac{2+8}{2} = 5$$, $$\beta = \frac{8-2}{2} = 3$$
• For $$(4,4)$$: $$\alpha = \frac{4+4}{2} = 4$$, $$\beta = \frac{4-4}{2} = 0$$
• For $$(8,2)$$: $$\alpha = \frac{8+2}{2} = 5$$, $$\beta = \frac{2-8}{2} = -3$$
• For $$(-2,-8)$$: $$\alpha = \frac{-2-8}{2} = -5$$, $$\beta = \frac{-8-(-2)}{2} = -3$$
• For $$(-4,-4)$$: $$\alpha = \frac{-4-4}{2} = -4$$, $$\beta = \frac{-4-(-4)}{2} = 0$$
• For $$(-8,-2)$$: $$\alpha = \frac{-8-2}{2} = -5$$, $$\beta = \frac{-2-(-8)}{2} = 3$$

The possible values of $$\alpha$$ are $$5, 4, -5, -4$$. Comparing with the options: A. $$3$$, B. $$5$$, C. $$17$$, D. $$9$$, the value $$\alpha = 5$$ is present (option B).

Verification for $$\alpha = 5$$, $$\beta = 3$$: $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & 3 & 5 \end{bmatrix}$$, $$|A| = 5^2 - 3^2 = 25 - 9 = 16$$. Then $$|2A| = 8 \times 16 = 128$$, and $$|2A|^3 = 128^3 = (2^7)^3 = 2^{21}$$, which satisfies the condition.

Checking other options:

• For $$\alpha = 3$$: $$\alpha^2 - \beta^2 = 9 - \beta^2 = 16 \implies \beta^2 = -7$$, not real.
• For $$\alpha = 17$$: $$289 - \beta^2 = 16 \implies \beta^2 = 273$$, not a perfect square.
• For $$\alpha = 9$$: $$81 - \beta^2 = 16 \implies \beta^2 = 65$$, not a perfect square.

Thus, the only valid value from the options is $$\alpha = 5$$.

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

 $$2A - A^T = \begin{bmatrix}1 & 3 & 2\alpha\\ 0 & 0 & 1\\ -\alpha & 1 & 2\end{bmatrix}$$

$$\det(2A - A^T) = 1(0-1) - 3(0+\alpha) + 2\alpha(0) = -1 - 3\alpha$$

Computing $$A - 2A^T = \begin{bmatrix}-1 & 0 & \alpha\\ -3 & 0 & -1\\ -2\alpha & -1 & -2\end{bmatrix}$$

$$\det(A - 2A^T) = -1(0-1) - 0 + \alpha(3-0) = 1 + 3\alpha$$

For a $$3 \times 3$$ matrix $$M$$: $$\det(\text{adj}(M)) = (\det M)^2$$. Therefore:

$$\det(\text{adj}(2A-A^T) \cdot \text{adj}(A-2A^T)) = (-1-3\alpha)^2 \cdot (1+3\alpha)^2 = (1+3\alpha)^4$$

Setting this equal to $$2^8 = 256$$:

$$(1+3\alpha)^4 = 256 \Rightarrow (1+3\alpha)^2 = 16 \Rightarrow 1+3\alpha = 4 \Rightarrow \alpha = 1$$

Therefore $$(\det A)^2 = (1 - 5)^2 = 16$$.

Matrix $$B$$ is stated to be the matrix of cofactors of the elements of $$A$$.
For any square matrix, if $$C$$ is its cofactor matrix, then

$$A\,C = \det(A)\,I_3 \qquad -(1)$$

because the scalar product of one row of $$A$$ with the cofactors of any different row vanishes, while the product with the cofactors of the same row equals $$\det(A)$$.

Taking determinant on both sides of $$(1)$$ gives

$$\det(A\,C)=\det(A)^3 \qquad -(2)$$

Since $$B$$ is that cofactor matrix, $$\det(AB)=\det(A)^3$$.
Thus our task reduces to finding $$\det(A)$$. To do this we first determine $$\alpha$$ and $$\beta$$ from the fact that the cofactors of $$A$$ equal the corresponding entries of $$B$$.

The needed cofactors of the first row of $$A=\begin{bmatrix}\beta & \alpha & 3\\ \alpha & \alpha & \beta\\ -\beta & \alpha & 2\alpha\end{bmatrix}$$ are computed below.

Case 1: Cofactor $$C_{11}$$ (remove row 1, column 1)

$$\begin{vmatrix}\alpha & \beta\\ \alpha & 2\alpha\end{vmatrix} = 2\alpha^2-\alpha\beta \quad\Longrightarrow\quad C_{11}=2\alpha^2-\alpha\beta$$

Given $$B_{11}=3\alpha$$, therefore

$$2\alpha^2-\alpha\beta = 3\alpha \quad\Longrightarrow\quad 2\alpha-\beta = 3 \qquad -(3)$$

Case 2: Cofactor $$C_{12}$$ (remove row 1, column 2)

Minor $$M_{12}= \begin{vmatrix}\alpha & \beta\\ -\beta & 2\alpha\end{vmatrix} =2\alpha^2+\beta^2$$
Cofactor $$C_{12}=(-1)^{1+2}M_{12}=-(2\alpha^2+\beta^2)$$

Given $$B_{12}=-9$$, therefore

$$-(2\alpha^2+\beta^2)=-9 \quad\Longrightarrow\quad 2\alpha^2+\beta^2 = 9 \qquad -(4)$$

Case 3: Cofactor $$C_{13}$$ (remove row 1, column 3)

$$\begin{vmatrix}\alpha & \alpha\\ -\beta & \alpha\end{vmatrix} =\alpha^2+\alpha\beta \quad\Longrightarrow\quad C_{13}= \alpha^2+\alpha\beta$$

Given $$B_{13}=3\alpha$$, therefore

$$\alpha^2+\alpha\beta = 3\alpha \quad\Longrightarrow\quad \alpha+\beta = 3 \qquad -(5)$$

Solving $$(3)$$ and $$(5)$$ simultaneously:

From $$(5)$$, $$\beta = 3-\alpha$$. Substituting in $$(3)$$:

$$2\alpha-(3-\alpha)=3 \quad\Longrightarrow\quad 3\alpha=6 \quad\Longrightarrow\quad \alpha=2$$

Then $$\beta = 3-\alpha = 1$$. These values also satisfy $$(4)$$ since $$2(2)^2+1^2 = 8+1 = 9$$.

Determinant of $$A$$ with $$\alpha=2,\;\beta=1$$

$$A=\begin{bmatrix}1 & 2 & 3\\ 2 & 2 & 1\\ -1 & 2 & 4\end{bmatrix}$$

Expanding along the first row,

$$\det(A)= 1\bigl(2\cdot4-1\cdot2\bigr) -2\bigl(2\cdot4-1\cdot(-1)\bigr) +3\bigl(2\cdot2-2\cdot(-1)\bigr)$$

$$\det(A)=1(8-2)-2(8+1)+3(4+2)=6-18+18=6 \qquad -(6)$$

Determinant of $$AB$$

Using $$(2)$$ and $$(6)$$,

$$\det(AB)=\det(A)^3 = 6^3 = 216$$

Therefore the correct option is Option B (216).

Given   $$\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

First operation will be $$C_2$$ $$\Rightarrow$$ $$C_2-C_3$$

$$\therefore$$ $$\begin{vmatrix} 1 & \frac{3}{2}- \alpha - \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3}-\alpha - \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\begin{vmatrix} 1 & - \alpha  & \alpha + \frac{3}{2} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1-0 & 0 \end{vmatrix} = 0$$

Now we will operate $$R_1$$ $$\Rightarrow$$ $$R_1-R_2$$

$$\therefore$$  $$\begin{vmatrix} 1-1 & \alpha- \alpha & \alpha + \frac{3}{2}-\alpha - \frac{1}{3} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\begin{vmatrix} 0 & 0 & \frac{7}{6} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\dfrac{7}{6}\left(\left(1\right)\left(3\alpha\ +1\right)-\left(-\alpha\ \right)\left(2\alpha\ +3\right)\right)=0$$

$$\therefore$$ $$2\alpha\ ^2+6\alpha\ +1=0$$

$$\therefore$$ $$\alpha\ =\dfrac{\left(-6\pm\ \sqrt{\ 28}\right)}{4}$$

$$\therefore$$ $$\alpha_1\ =-1.5+\dfrac{\sqrt{\ 7}}{2}$$ and $$\alpha_2\ =-1.5-\dfrac{\sqrt{\ 7}}{2}$$

$$\sqrt{\ 7}\simeq\ 2.645$$

$$\therefore$$ $$\alpha\ _1=-0.177$$ and $$\alpha\ _2=-2.8228$$

$$\therefore$$ $$\alpha\ \in\ \left(0,-3\right)$$

Hence, correct option is B.

First, compute $$A^T$$, the transpose of A:

$$A^T = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix}$$.

Now, compute $$A^T A$$:

$$A^T A = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$$.

Thus, $$A^T A = 3I$$.

 $$C = ABA^T$$.

 $$C^2 = (ABA^T)(ABA^T) = AB(A^T A)BA^T = AB(3I)BA^T = 3AB^2A^T$$, since scalar multiplication commutes.

 compute $$B^2$$:

$$B^2 = B \cdot B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} (1)(1) + (0)(1) & (1)(0) + (0)(1) \\ (1)(1) + (1)(1) & (1)(0) + (1)(1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$$.

So, $$C^2 = 3A B^2 A^T = 3A \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} A^T$$.

Now, $$X = A^T C^2 A = A^T \left(3A B^2 A^T\right) A = 3 A^T A B^2 A^T A$$.

Substitute $$A^T A = 3I$$:

$$X = 3 (3I) B^2 (3I) = 3 \cdot 3 \cdot 3 \cdot I B^2 I = 27 B^2$$, since the identity matrix commutes.

$$X = 27 \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 27 & 0 \\ 54 & 27 \end{pmatrix}$$.

compute $$\det X$$:

For a matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, $$\det = ad - bc$$.

So, $$\det X = (27)(27) - (0)(54) = 729 - 0 = 729$$.

$$|\text{adj}(kM)| = k^{n(n-1)} |\text{adj}(M)| = k^{6}|M|^2$$ for order $$3$$, and $$|\text{adj}(M)| = |M|^2$$.

Inner term: $$|(2A)^{-1}| = \frac{1}{8|A|} = \frac{1}{24}$$.

$$|3\text{adj}((2A)^{-1})| = 3^3 \cdot \left(\frac{1}{24}\right)^2 = \frac{27}{576} = \frac{3}{64}$$.

$$|-3\text{adj}(\dots)| = (-3)^3 \cdot \left(\frac{3}{64}\right)^2 = -\frac{243}{4096}$$.

$$|-4\text{adj}(\dots)| = (-4)^3 \cdot \left(-\frac{243}{4096}\right)^2 = -64 \cdot \frac{59049}{16777216} = -\frac{59049}{262144}$$.

Final determinant is the square of the previous result: $$\left(-\frac{59049}{262144}\right)^2 = \frac{3^{20}}{2^{36}} = 2^{-36} \cdot 3^{20}$$.

Matching parameters: $$m = -36$$, $$n = 20 \implies m + 2n = -36 + 40 = \mathbf{4}$$

Given the determinant function $$f(x) = \begin{vmatrix} 2\cos^4 x & 2\sin^4 x & 3 + \sin^2 2x \\ 3 + 2\cos^4 x & 2\sin^4 x & \sin^2 2x \\ 2\cos^4 x & 3 + 2\sin^4 x & \sin^2 2x \end{vmatrix}$$, find $$\frac{1}{5}f'(0)$$.

Let $$a = 2\cos^4 x$$, $$b = 2\sin^4 x$$, $$c = \sin^2 2x$$.

Note that $$a + b + c = 2\cos^4 x + 2\sin^4 x + \sin^2 2x = 2(\cos^4 x + \sin^4 x) + 4\sin^2 x\cos^2 x$$

$$= 2(\cos^2 x + \sin^2 x)^2 - 4\sin^2 x\cos^2 x + 4\sin^2 x\cos^2 x = 2$$.

So $$a + b + c = 2$$ and the third column of row 1 is $$3 + c = 3 + c$$, while the first column entries are $$a, 3+a, a$$.

Apply $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} a & b & 3+c \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix}$$

Expanding: $$a(0-(-9)) - b(-9-0) + (3+c)(9-0)$$

$$= 9a + 9b + 9(3+c) = 9(a + b + 3 + c) = 9(2 + 3) = 45$$.

So $$f(x) = 45$$ for all $$x$$ (constant function).

Since $$f(x) = 45$$ is constant, $$f'(x) = 0$$ for all $$x$$.

$$\frac{1}{5}f'(0) = \frac{0}{5} = 0$$.

The correct answer is Option A: 0.

A homogeneous system $$AX = 0$$ has a non-trivial solution if and only if the determinant of the coefficient matrix is zero ($$|A| = 0$$).

Set up the determinant:

$$\begin{vmatrix} 1 & \sqrt{2}\sin\alpha & \sqrt{2}\cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ 1 & \sin\alpha & -\cos\alpha \end{vmatrix} = 0$$

Perform row operations ($$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$) or expand directly. Expanding along the first column:

$$1(-\cos^2\alpha - \sin^2\alpha) - 1(-\sqrt{2}\sin\alpha\cos\alpha - \sqrt{2}\sin\alpha\cos\alpha) + 1(\sqrt{2}\sin^2\alpha - \sqrt{2}\cos^2\alpha) = 0$$

Simplify:

$$-1 + 2\sqrt{2}\sin\alpha\cos\alpha - \sqrt{2}(\cos^2\alpha - \sin^2\alpha) = 0$$

$$-1 + \sqrt{2}\sin 2\alpha - \sqrt{2}\cos 2\alpha = 0 \implies \sin 2\alpha - \cos 2\alpha = \frac{1}{\sqrt{2}}$$

Divide by $$\sqrt{2}$$: $$\frac{1}{\sqrt{2}}\sin 2\alpha - \frac{1}{\sqrt{2}}\cos 2\alpha = \frac{1}{2} \implies \sin(2\alpha - \frac{\pi}{4}) = \frac{1}{2}$$

$$2\alpha - \frac{\pi}{4} = \frac{\pi}{6} \implies 2\alpha = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \implies \alpha = \frac{5\pi}{24}$$

$$A$$ 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 system is: $$x + y + z = 4\mu$$, $$x + 2y + 2\lambda z = 10\mu$$, $$x + 3y + 4\lambda^2 z = \mu^2 + 15$$.

The coefficient matrix determinant: $$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$$

$$= 1(8\lambda^2 - 6\lambda) - 1(4\lambda^2 - 2\lambda) + 1(3 - 2) = 8\lambda^2 - 6\lambda - 4\lambda^2 + 2\lambda + 1 = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$$

$$D = 0$$ when $$\lambda = \frac{1}{2}$$.

When $$\lambda = \frac{1}{2}$$: the system becomes $$x + y + z = 4\mu$$, $$x + 2y + z = 10\mu$$, $$x + 3y + z = \mu^2 + 15$$.

From equations 1 and 2: $$y = 6\mu$$. From equations 2 and 3: $$y = \mu^2 + 15 - 10\mu$$.

So $$6\mu = \mu^2 - 10\mu + 15 \Rightarrow \mu^2 - 16\mu + 15 = 0 \Rightarrow (\mu-1)(\mu-15) = 0$$.

For consistency at $$\lambda = \frac{1}{2}$$: $$\mu = 1$$ or $$\mu = 15$$.

Checking the options:

Option (1): Unique solution if $$\lambda \neq \frac{1}{2}$$ and $$\mu \neq 1, 15$$ — True (since $$D \neq 0$$).

Option (2): Inconsistent if $$\lambda = \frac{1}{2}$$ and $$\mu \neq 1$$ — This should say $$\mu \neq 1$$ AND $$\mu \neq 15$$. If $$\mu = 15$$, the system is consistent. So if $$\mu \neq 1$$ but $$\mu = 15$$, it's consistent. Hence this statement is NOT correct.

Option (3): Infinite solutions if $$\lambda = \frac{1}{2}$$ and $$\mu = 15$$ — True.

Option (4): Consistent if $$\lambda \neq \frac{1}{2}$$ — True (unique solution exists).

The answer is Option (2): The statement that is NOT correct.

Since the determinant is linear in Column 1, we can bring the sum inside:

$$\sum_{r=1}^n A_r = \begin{vmatrix} \sum r & 1 & \frac{n^2}{2} + \alpha \\ \sum 2r & 2 & n^2 - \beta \\ \sum (3r - 2) & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$$

Using standard AP sum formulas for $$r = 1 \text{ to } n$$:

$$\sum r = \frac{n^2+n}{2}, \quad \sum 2r = n^2+n, \quad \sum (3r-2) = \frac{n(3n-1)}{2}$$

Substitute $$n = 1$$ as a shortcut:

$$\det = \begin{vmatrix} 1 & 1 & \frac{1}{2}+\alpha \\ 2 & 2 & 1-\beta \\ 1 & 3 & 1 \end{vmatrix}$$

Apply row operation $$R_2 \to R_2 - 2R_1$$:

$$\det = \begin{vmatrix} 1 & 1 & \frac{1}{2}+\alpha \\ 0 & 0 & -2\alpha-\beta \\ 1 & 3 & 1 \end{vmatrix} = -(-2\alpha - \beta)(3 - 1) = 4\alpha + 2\beta$$

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 $$\boxed{10}$$.

Since $$\text{adj}(A) = \det(A)\cdot A^{-1}$$ when $$A$$ is invertible

taking determinants on both sides gives $$\det(\text{adj}(A)) = (\det(A))^n \cdot \det(A^{-1}) = (\det(A))^n \cdot \frac{1}{\det(A)} = (\det(A))^{n-1}.$$

For $$n = 3$$, this becomes $$\det(\text{adj}(A)) = (\det(A))^2.$$

 Let $$D_k$$ be the determinant after applying the adjoint operation $$k$$ times.

We have $$D_0 = \det(A) = 2,$$ and each adjoint operation squares the determinant, so $$D_1 = D_0^2 = 2^2 = 4,$$ $$D_2 = D_1^2 = 4^2 = 2^4,$$ $$D_3 = D_2^2 = (2^4)^2 = 2^8.$$ In general, one finds $$D_k = 2^{2^k}.$$

 $$D_{2024} \bmod 9$$ = $$2^{2^{2024}} \bmod 9.$$

To proceed, we note the pattern of powers of 2 modulo 9: $$2^1 = 2,\quad 2^2 = 4,\quad 2^3 = 8,\quad 2^4 = 16 \equiv 7,\quad 2^5 = 32 \equiv 5,\quad 2^6 = 64 \equiv 1 \pmod{9}.$$ Thus the cycle length is 6, so the value of $$2^m \bmod 9$$ depends only on $$m \bmod 6$$.

We next determine $$2^{2024} \bmod 6$$ by computing it modulo 2 and modulo 3 separately

$$2^{2024} \bmod 2 = 0.$$

Moreover, because $$2 \equiv -1 \pmod{3},$$ $$2^{2024} \equiv (-1)^{2024} = 1 \pmod{3}.$$ 

By the Chinese Remainder Theorem, the unique solution modulo 6 satisfying these conditions is $$2^{2024} \equiv 4 \pmod{6}.$$

It follows that $$2^{2^{2024}} \equiv 2^4 = 16 \equiv 7 \pmod{9}.$$

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 $$a_{ij} = \begin{cases} 0 & i \neq j \\ 3 & i = j \end{cases}$$

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

For an $$n \times n$$ matrix, $$\det(kA) = k^n \det(A)$$ and $$\det(\text{adj}(A)) = (\det A)^{n-1}$$. Here, $$n=3$$.

The expression is $$\det(3 \text{ adj}(2 \text{ adj}(d A)))$$.

$$\det(\text{adj}(dA)) = (d \cdot d^3)^2 = d^8$$.

$$\det(2 \text{ adj}(dA)) = 2^3 \cdot d^8$$.

$$\det(\text{adj}(2 \text{ adj}(dA))) = (2^3 d^8)^2 = 2^6 d^{16}$$.

$$\det(3 \text{ adj}(\dots)) = 3^3 \cdot 2^6 \cdot d^{16}$$.

Set $$3^3 \cdot 2^6 \cdot d^{16} = 3^{-13} \cdot 2^{-10} \implies d^{16} = 3^{-16} \cdot 2^{-16} \implies d = \frac{1}{6}$$.

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

Substitute $$d = 2^{-1} \cdot 3^{-1}$$:

$$3^3 \cdot 2^6 \cdot (2^{-2} \cdot 3^{-2}) = 2^4 \cdot 3^1$$.

So, $$m = 4, n = 1$$.

$$|3(4) + 2(1)| = |12 + 2| = \mathbf{14}$$

We begin by finding the characteristic equation of the matrix A. Computing $$\det(A - \lambda I) = (2-\lambda)(1-\lambda) + 1 = \lambda^2 - 3\lambda + 3 = 0$$ and applying Cayley-Hamilton yields $$A^2 = 3A - 3I$$.

Next, letting $$t_n = \mathrm{tr}(A^n)$$, the characteristic equation implies the recurrence $$t_n = 3\,t_{n-1} - 3\,t_{n-2}$$ with initial values $$t_0 = \mathrm{tr}(I) = 2$$ and $$t_1 = \mathrm{tr}(A) = 3$$.

Using this recurrence, we compute $$t_2 = 3(3) - 3(2) = 3$$, $$t_3 = 3(3) - 3(3) = 0$$, $$t_4 = 3(0) - 3(3) = -9$$, $$t_5 = 3(-9) - 3(0) = -27$$, $$t_6 = 3(-27) - 3(-9) = -54$$ and $$t_7 = 3(-54) - 3(-27) = -81$$.

Alternatively, the eigenvalues of A are $$\lambda = \frac{3 \pm i\sqrt{3}}{2} = \sqrt{3}\,e^{\pm i\pi/6}$$, so that $$\lambda^n = 3^{n/2}\,e^{\pm in\pi/6}$$ and therefore $$t_n = \lambda_1^n + \lambda_2^n = 2 \cdot 3^{n/2} \cos\frac{n\pi}{6}$$.

For $$n = 13$$ this yields $$t_{13} = 2 \cdot 3^{13/2} \cos\frac{13\pi}{6} = 2 \cdot 3^{13/2} \cos\frac{\pi}{6} = 2 \cdot 3^{13/2} \cdot \frac{\sqrt{3}}{2} = 3^{13/2} \cdot 3^{1/2} = 3^7$$ and hence $$n = 7$$.

Therefore, the final answer is 7.

$$x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0)$$ has nontrivial solution, so determinant = 0:

$$\begin{vmatrix}\alpha&1&2\\1&\beta&2\\2&3&\gamma\end{vmatrix} = 0$$.

$$\alpha(\beta\gamma-6)-1(\gamma-4)+2(3-2\beta) = 0$$.

$$\alpha\beta\gamma - 6\alpha - \gamma + 4 + 6 - 4\beta = 0$$.

Since $$\alpha\beta\gamma = 45$$: $$45 - 6\alpha - \gamma - 4\beta + 10 = 0 \Rightarrow 6\alpha + 4\beta + \gamma = 55$$.

The answer is 55.

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

Write the three equations in the standard form

$$\begin{aligned} x+2y+z &= 7 \hspace{20pt} -(1)\\ x+\alpha z &= 11 \hspace{20pt} -(2)\\ 2x-3y+\beta z &= \gamma \hspace{20pt} -(3) \end{aligned}$$

The coefficient matrix is

$$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & 0 & \alpha\\ 2 & -3 & \beta \end{bmatrix}$$

The determinant of $$A$$ decides whether a unique solution exists.

$$\begin{aligned} \det(A) &= \begin{vmatrix} 1 & 2 & 1\\ 1 & 0 & \alpha\\ 2 & -3 & \beta \end{vmatrix}\\[2pt] &=1\,(0\cdot\beta-\alpha(-3))- 2\,(1\cdot\beta-\alpha\cdot2)+ 1\,(1\cdot(-3)-0\cdot2)\\[2pt] &=1\,(3\alpha)-2\,(\beta-2\alpha)+(-3)\\[2pt] &=3\alpha-2\beta+4\alpha-3\\ &=7\alpha-2\beta-3 \end{aligned}$$

Thus

$$\det(A)=0 \;\Longleftrightarrow\; 2\beta=7\alpha-3 \quad -(4)$$

When $$\det(A)\neq0$$ the system has a unique solution; when $$\det(A)=0$$ we must compare the ranks of $$A$$ and the augmented matrix $$[A|B]$$ to decide between “no solution” and “infinitely many solutions”.

Next, express row (3) of $$[A|B]$$ as a linear combination of rows (1) and (2).
Take numbers $$p,q$$ so that

$$p(1,2,1)+q(1,0,\alpha)=(2,-3,\beta)$$

Equating components gives

$$p+q=2,\;2p=-3\;\Longrightarrow\;p=-\tfrac32,\;q=\tfrac72$$

With this $$z$$-coefficient becomes

$$p\cdot1+q\cdot\alpha=-\tfrac32+\tfrac72\alpha=\beta$$

which is exactly condition (4). The corresponding constant term is

$$p\cdot7+q\cdot11=-\tfrac32\cdot7+\tfrac72\cdot11 =-\tfrac{21}{2}+\tfrac{77}{2}=28$$

Hence, when (4) holds:
  • If $$\gamma=28$$, row (3) is a linear combination of rows (1) and (2) ⇒ ranks are equal (both 2 < 3) ⇒ infinitely many solutions.
  • If $$\gamma\neq28$$, the augmented matrix has rank 3 while $$\operatorname{rank}(A)=2$$ ⇒ no solution.

Now fix $$\alpha=1$$ as given in cases (R) and (S). From (4) we get the special value

$$2\beta=7(1)-3\;\Longrightarrow\;\beta=2$$

Therefore, for $$\alpha=1$$:

$$\det(A)=7(1)-2\beta-3=4-2\beta$$

If $$\beta\neq2$$ then $$\det(A)\neq0$$, so the system has a unique solution irrespective of $$\gamma$$. In particular, substituting $$\alpha=1$$ and solving quickly:

From (2): $$x=11-z$$. Substituting into (1): $$11-z+2y+z=7\Rightarrow y=-2$$.
Using (3): $$2(11-z)-3(-2)+\beta z=\gamma$$ gives $$28+(\beta-2)z=\gamma$$.

• For $$\gamma=28$$ we can take $$z=0$$, leading to the specific solution $$x=11,\;y=-2,\;z=0$$ (unique because $$\beta\neq2$$ keeps $$\det(A)\neq0$$).
• For $$\gamma\neq28$$, solving $$z=\dfrac{\gamma-28}{\beta-2}$$ yields a single real triple, again a unique solution.

We are now ready to match.

Case P: $$\beta=\tfrac12(7\alpha-3),\;\gamma=28$$ ⇒ $$\det(A)=0$$ and ranks equal ⇒ infinitely many solutions ⇒ List-II (3).

Case Q: $$\beta=\tfrac12(7\alpha-3),\;\gamma\neq28$$ ⇒ $$\det(A)=0$$ but ranks unequal ⇒ no solution ⇒ List-II (2).

Case R: $$\alpha=1,\;\beta\neq2,\;\gamma\neq28$$ ⇒ $$\det(A)\neq0$$ ⇒ unique solution ⇒ List-II (1).

Case S: $$\alpha=1,\;\beta\neq2,\;\gamma=28$$ gives the concrete solution $$x=11,\;y=-2,\;z=0$$ (and the solution is unique) ⇒ List-II (4).

The only option that lists these four pairings is:

Option A which is: (P) → (3), (Q) → (2), (R) → (1), (S) → (4).

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: $$-16$$.

Given the system:

$$2x + 4y + 2az = b \quad \cdots(1)$$

$$x + 2y + 3z = 4 \quad \cdots(2)$$

$$2x + 5y + 2z = 8 \quad \cdots(3)$$

Determinant of coefficient matrix:

$$\Delta = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & 5 & 2 \end{vmatrix} = 2(4-15) - 4(2-6) + 2a(5-4) = -22 + 16 + 2a = 2a - 6$$

$$\Delta = 0$$ when $$a = 3$$.

For $$a \neq 3$$: System has a unique solution regardless of $$b$$.

For $$a = 3$$: Equation (1) becomes $$2x + 4y + 6z = b$$, i.e., $$2(x + 2y + 3z) = b$$.

From (2): $$x + 2y + 3z = 4$$, so (1) requires $$b = 8$$.

Checking each option:

Option A: $$a = b = 6$$: $$\Delta = 6 \neq 0$$. Unique solution. ✓ (Correct statement)

Option B: $$a = 3, b = 6$$: $$\Delta = 0$$, but (1) gives $$8 = 6$$, contradiction. No solution, NOT infinitely many. ✗ (Incorrect statement)

Option C: $$a = 3, b = 8$$: $$\Delta = 0$$, consistent. Infinitely many solutions. ✓

Option D: $$a = b = 8$$: $$\Delta = 10 \neq 0$$. Unique solution. ✓

The statement that is NOT correct is Option B.

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 need to find $$|3 \cdot \text{adj}(|3A| \cdot A^2)|$$ for a $$3 \times 3$$ matrix $$A$$ with $$|A| = 2$$.

First, $$|3A| = 3^3 |A| = 27 \times 2 = 54$$. So the matrix inside the adjoint is $$54 \cdot A^2$$, which is a scalar times a matrix.

Now, $$|54 A^2| = 54^3 \cdot |A|^2 = 54^3 \times 4$$. For a $$3 \times 3$$ matrix $$M$$, $$|\text{adj}(M)| = |M|^{n-1} = |M|^2$$. Therefore $$|\text{adj}(54A^2)| = |54A^2|^2 = (54^3 \times 4)^2 = 54^6 \times 16$$.

Finally, $$|3 \cdot \text{adj}(54A^2)| = 3^3 \cdot |\text{adj}(54A^2)| = 27 \times 54^6 \times 16$$.

Simplifying: $$54 = 2 \times 27 = 2 \times 3^3$$, so $$54^6 = 2^6 \times 3^{18}$$. Then $$27 \times 54^6 \times 16 = 3^3 \times 2^6 \times 3^{18} \times 2^4 = 3^{21} \times 2^{10}$$.

We can write this as $$3^{11} \times 3^{10} \times 2^{10} = 3^{11} \times 6^{10}$$.

The correct answer is Option D: $$3^{11} \cdot 6^{10}$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

Given $$A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$$ with $$x, y, z > 1$$.

Finding $$|A|$$:

Let $$p = \ln x, q = \ln y, r = \ln z$$. Multiply rows 1, 2, 3 by $$p, q, r$$ respectively:

$$M = \begin{bmatrix} p & q & r \\ p & 2q & r \\ p & q & 3r \end{bmatrix}$$

$$\det(M) = pqr \cdot \det(A)$$

Applying $$R_2 - R_1$$ and $$R_3 - R_1$$:

$$\det(M) = \begin{vmatrix} p & q & r \\ 0 & q & 0 \\ 0 & 0 & 2r \end{vmatrix} = 2pqr$$

Therefore $$\det(A) = 2$$.

Verification: Setting $$x = y = z$$: $$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{bmatrix}$$, $$\det = 5 - 2 - 1 = 2$$ ✓

Computing $$|\text{adj}(\text{adj}(A^2))|$$:

For an $$n \times n$$ matrix $$M$$: $$|\text{adj}(M)| = |M|^{n-1}$$

$$|\text{adj}(\text{adj}(M))| = |M|^{(n-1)^2}$$

With $$n = 3$$ and $$M = A^2$$:

$$|A^2| = |A|^2 = 4$$

$$|\text{adj}(\text{adj}(A^2))| = |A^2|^{(3-1)^2} = 4^4 = 256 = 2^8$$

The answer is Option B: $$2^8$$.

We need to find all $$t \in \mathbb{R}$$ for which the given matrix is invertible (determinant $$\neq 0$$).

By factoring out $$e^t$$ from column 1, $$e^{-t}$$ from column 2, and $$e^{-t}$$ from column 3, the determinant becomes $$\det = e^{-t} \cdot \det \begin{bmatrix} 1 & \sin t - 2\cos t & -2\sin t - \cos t \\[6pt] 1 & 2\sin t + \cos t & \sin t - 2\cos t \\[6pt] 1 & \cos t & \sin t \end{bmatrix}$$.

Applying the row operations $$R_1 \to R_1 - R_3$$ and $$R_2 \to R_2 - R_3$$ gives Row 1 as $$(0,\ \sin t - 2\cos t - \cos t,\ -2\sin t - \cos t - \sin t) = (0,\ \sin t - 3\cos t,\ -3\sin t - \cos t)$$ and Row 2 as $$(0,\ 2\sin t + \cos t - \cos t,\ \sin t - 2\cos t - \sin t) = (0,\ 2\sin t,\ -2\cos t)$$. Therefore the determinant is $$e^{-t} \det \begin{bmatrix} 0 & \sin t - 3\cos t & -3\sin t - \cos t \\[6pt] 0 & 2\sin t & -2\cos t \\[6pt] 1 & \cos t & \sin t \end{bmatrix}$$.

Expanding along the first column yields $$e^{-t}\Bigl[(\sin t - 3\cos t)(-2\cos t) - (-3\sin t - \cos t)(2\sin t)\Bigr],$$ which simplifies to $$e^{-t}\Bigl[-2\sin t\cos t + 6\cos^2 t + 6\sin^2 t + 2\sin t\cos t\Bigr] = e^{-t}\bigl[6\cos^2 t + 6\sin^2 t\bigr] = 6e^{-t}.$$

Since $$6e^{-t} \neq 0$$ for all $$t \in \mathbb{R}$$, the determinant never vanishes, and the matrix is invertible for all real $$t$$. Hence the correct answer is Option D: $$\mathbb{R}$$.

The system of equations is $$2x - y + 3z = 5$$
$$3x + 2y - z = 7$$
$$4x + 5y + \alpha z = \beta$$.

The determinant of the coefficient matrix is $$D = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{vmatrix}$$. Expanding along the first row gives $$D = 2(2\alpha + 5) + 1(3\alpha + 4) + 3(15 - 8) = 4\alpha + 10 + 3\alpha + 4 + 21 = 7\alpha + 35 = 7(\alpha + 5)$$.

This determinant is zero exactly when $$\alpha = -5$$. For $$\alpha \neq -5$$ the determinant is nonzero, so the system has a unique solution in that case.

When $$\alpha = -5$$ and there are infinitely many solutions, the third equation must be a linear combination of the first two: $$(3) = p\,(1) + q\,(2)$$. Matching coefficients yields
$$2p + 3q = 4$$
$$-p + 2q = 5$$
$$3p - q = -5$$. Solving the second and third equations gives $$p = -1$$ and $$q = 2$$. Substituting these into the constant terms gives $$\beta = p(5) + q(7) = -5 + 14 = 9$$.

Now each option can be checked. Option (1) asserts infinitely many solutions when $$\alpha = -5$$ and $$\beta = 9$$, which agrees with the result above.

Option (2) proposes infinitely many solutions for $$\alpha = -6$$ and $$\beta = 9$$. However, when $$\alpha = -6$$ the determinant is $$D = 7(-6 + 5) = -7 \neq 0$$, so the system actually has a unique solution rather than infinitely many.

Option (3) states the system is inconsistent for $$\alpha = -5$$ and $$\beta = 8$$. In this case $$D = 0$$ but $$\beta \neq 9$$, so the third equation does not align with the first two and the system is indeed inconsistent.

Option (4) claims a unique solution for any $$\alpha \neq -5$$ (regardless of $$\beta$$), which is correct because the determinant is nonzero whenever $$\alpha \neq -5$$.

Therefore the statement that is not correct is Option (2).

We need to find which statement is NOT correct about the system: $$\alpha x + y + z = 1$$, $$x + \alpha y + z = 1$$, $$x + y + \alpha z = \beta$$.

First we write the coefficient matrix and find its determinant: $$D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix}$$. Expanding gives $$D = \alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha)$$, which simplifies to $$\alpha^3 - \alpha - \alpha + 1 + 1 - \alpha$$, then to $$\alpha^3 - 3\alpha + 2$$, and finally to $$(\alpha - 1)^2(\alpha + 2)$$.

Option A: $$\alpha = 2, \beta = -1$$ -- infinitely many solutions. When $$\alpha = 2$$, the determinant becomes $$(2-1)^2(2+2) = 4 \neq 0$$, so the system has a unique solution. Thus the claim of infinitely many solutions is NOT correct.

Option B: $$\alpha = -2, \beta = 1$$ -- no solution. When $$\alpha = -2$$, the determinant is $$(-3)^2(0) = 0$$, so the system is singular. Adding all three equations: $$(-2+1+1)x + (1-2+1)y + (1+1-2)z = 1 + 1 + \beta$$ gives $$0 = 2 + \beta$$, and with $$\beta = 1$$ this leads to $$0 = 3$$, a contradiction. Therefore there is no solution, confirming the claim.

Option C: $$\alpha = 2, \beta = 1$$ -- $$x + y + z = 3/4$$. When $$\alpha = 2, \beta = 1$$ the determinant is $$4 \neq 0$$ so a unique solution exists. The system becomes $$2x + y + z = 1$$, $$x + 2y + z = 1$$, $$x + y + 2z = 1$$. Adding these yields $$4(x+y+z) = 3$$, giving $$x+y+z = 3/4$$, which verifies the claim.

Option D: $$\alpha = 1, \beta = 1$$ -- infinitely many solutions. When $$\alpha = 1$$, the determinant is $$0$$ and all three equations reduce to $$x + y + z = 1$$. With $$\beta = 1$$ all three are identical, leading to infinitely many solutions, so this claim is correct.

The statement that is NOT correct is Option A.

Given system:

$$x + y + z = 6$$ ... (i)

$$\alpha x + \beta y + 7z = 3$$ ... (ii)

$$x + 2y + 3z = 14$$ ... (iii)

(iii) - (i): $$y + 2z = 8 \Rightarrow y = 8 - 2z$$ ... (iv)

From (i): $$x = 6 - y - z = 6 - (8-2z) - z = z - 2$$ ... (v)

$$\alpha(z-2) + \beta(8-2z) + 7z = 3$$

$$z(\alpha - 2\beta + 7) + (-2\alpha + 8\beta) = 3$$

For infinitely many solutions, this must be true for all $$z$$:

$$\alpha - 2\beta + 7 = 0$$ and $$-2\alpha + 8\beta = 3$$

From first: $$\alpha = 2\beta - 7$$. Substituting: $$-2(2\beta - 7) + 8\beta = 3 \Rightarrow -4\beta + 14 + 8\beta = 3 \Rightarrow 4\beta = -11 \Rightarrow \beta = -\frac{11}{4}$$

$$\alpha = 2(-\frac{11}{4}) - 7 = -\frac{11}{2} - 7 = -\frac{25}{2}$$

Check if $$(\alpha, \beta) = (-\frac{25}{2}, -\frac{11}{4})$$ is on $$x + 2y + 18 = 0$$:

$$-\frac{25}{2} + 2(-\frac{11}{4}) + 18 = -\frac{25}{2} - \frac{11}{2} + 18 = -18 + 18 = 0$$ ✓

Option (1): $$\alpha = \beta = 7$$, no solution

$$z(7 - 14 + 7) + (-14 + 56) = 0 + 42 = 42 \neq 3$$. Inconsistent, so no solution. ✓ TRUE.

Option (2): $$\alpha = \beta, \alpha \neq 7$$, unique solution

$$z(\alpha - 2\alpha + 7) + (-2\alpha + 8\alpha) = 3$$

$$z(7 - \alpha) + 6\alpha = 3$$

Since $$\alpha \neq 7$$, $$7 - \alpha \neq 0$$, so $$z = \frac{3 - 6\alpha}{7 - \alpha}$$, unique. ✓ TRUE.

Option (3): Unique $$(\alpha, \beta)$$ on $$x + 2y + 18 = 0$$ for infinitely many solutions

We found a unique such point $$(-25/2, -11/4)$$. ✓ TRUE.

Option (4): For every $$(\alpha, \beta) \neq (7,7)$$ on $$x - 2y + 7 = 0$$, infinitely many solutions

On $$\alpha - 2\beta + 7 = 0$$, so the coefficient of $$z$$ vanishes. Then $$-2\alpha + 8\beta = 3$$ must hold.

$$\alpha = 2\beta - 7$$, so $$-2(2\beta - 7) + 8\beta = -4\beta + 14 + 8\beta = 4\beta + 14 = 3 \Rightarrow \beta = -11/4$$.

So only $$\beta = -11/4$$ works, not "every point" on the line. For other points on $$x - 2y + 7 = 0$$, we get $$0 \cdot z + \text{(nonzero)} = 3$$... wait, we get $$-2\alpha + 8\beta \neq 3$$, so NO solution (not infinitely many).

So Option (4) is NOT true. ✓

The correct answer is Option (4): $$\boxed{$$ For every point $$(\alpha,\beta) \neq (7,7)$$ on $$x - 2y + 7 = 0,$$ the system has infinitely many solutions. $$}$$

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 $$|\text{Adj } P| = 1$$.