Question

Let \[ A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix} \text{be a singular matrix for} \]

• A. \( k = 2 \) only
• B. \( k = \pm 2 \) only
• C. no real value of \( k \)
• D. all real values of \( k \)

Hint:

A matrix is singular if and only if its determinant is zero. For 3x3 matrices, expanding along a row or column with a zero simplifies the calculation. Always look for such shortcuts in determinant evaluation.

Answer 1

The Correct Option is D

A matrix is singular if its determinant is zero. We compute the determinant of \( A \): \[ \text{det}(A) = \begin{vmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{vmatrix} \] Expand along the first row: \[ = 0 \cdot \begin{vmatrix} -4 & -6 \\ -3 & -5 \end{vmatrix} - k \cdot \begin{vmatrix} k & -6 \\ k & -5 \end{vmatrix} + k \cdot \begin{vmatrix} k & -4 \\ k & -3 \end{vmatrix} \] Now compute the minors: \[ \begin{vmatrix} k & -6 \\ k & -5 \end{vmatrix} = k(-5) - (-6)(k) = -5k + 6k = k \] \[ \begin{vmatrix} k & -4 \\ k & -3 \end{vmatrix} = k(-3) - (-4)(k) = -3k + 4k = k \] Substitute back: \[ \text{det}(A) = -k(k) + k(k) = -k^2 + k^2 = 0 \] So for all real values of \( k \), the determinant is zero. Hence, \( A \) is singular for all real values of \( k \). \[ \boxed{\text{All real values of } k} \]