<p>The number of values of <i>x</i> for which the matrix \(A = \begin{pmatrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \end{pmatrix}\) is singular, is</p>
Step-by-Step Solution
Key Concept: A matrix is singular when its determinant equals zero. Set det(A) = 0 and solve the resulting polynomial equation.
<p><strong>Step 1:</strong> A matrix is singular when det(<i>A</i>) = 0.</p><p><strong>Step 2:</strong> Compute the determinant by expanding along the first row:</p><p>$\det(A) = (3-x)\begin{vmatrix} 4-x & 1 \\ -4 & -1-x \end{vmatrix} - 2\begin{vmatrix} 2 & 1 \\ -2 & -1-x \end{vmatrix} + 2\begin{vmatrix} 2 & 4-x \\ -2 & -4 \end{vmatrix}$</p><p><strong>Step 3:</strong> After calculation, this yields a cubic equation in <i>x</i>: $x^3 - 6x^2 + 11x - 6 = 0$, which factors as $(x-1)(x-2)(x-3) = 0$.</p><p><strong>Step 4:</strong> Therefore, <i>x</i> = 1, 2, or 3 makes the matrix singular (3 values).</p><p>∴ Answer is (d).</p>
Correct Answer: D