Question:

<p><strong>Step 1:</strong> Construct matrix $A$ explicitly:</p><p>$$A = \begin{bmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{bmatrix}$$</p><p><strong>Step 2:</strong> Calculate $\det(A)$ using cofactor expansion along row 1:</p><p>$$\det(A) = x(x^2 - 1) - 1(x - 0) + 0 = x^3 - x - x = x^3 - 2x = x(x^2 - 2)$$</p><p><strong>Step 3:</strong> Analyze key properties:</p><p>• <strong>Trace$(A)$</strong> $= x + x + x = 3x$ (sum of eigenvalues)</p><p>• <strong>$\det(A)$</strong> $= x(x^2 - 2) = x^3 - 2x$</p><p>• Matrix is <strong>symmetric</strong> ($a_{ij} = a_{ji}$), so all eigenvalues are real</p><p>• Sum of products of eigenvalues taken two at a time $=$ sum of all $2 \times 2$ principal minors $= (x^2 - 1) + (x^2 - 1) + (x^2 - 1) = 3x^2 - 3$</p><p><strong>Step 4:</strong> Verify statements:</p><p>• <strong>Option 1:</strong> For $x=2$, $A$ is a diagonal matrix: <strong>False</strong> (off-diagonal entries are 1)</p><p>• <strong>Option 2:</strong> $A$ is a symmetric matrix: <strong>True</strong> (for all $x$)</p><p>• <strong>Option 3:</strong> For $x=2$, $\det(A) = 2(4-2) = 2(2) = 4 \neq 6$: <strong>False</strong></p><p>• <strong>Option 4:</strong> Let $f(x) = \det(A) = x^3 - 2x$. Then $f'(x) = 3x^2 - 2 = 0$ gives $x = \pm\sqrt{\frac{2}{3}}$. Since $f''(x) = 6x$, we have a local maximum at $x = -\sqrt{\frac{2}{3}}$ and a local minimum at $x = \sqrt{\frac{2}{3}}$: <strong>True</strong></p><p>∴ Answer: <strong>B, D</strong></p>