The value of k for which the matrix \(\begin{bmatrix}2 & k\\ k^{2} & 4\end{bmatrix}\) is a singular matrix is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
A matrix is singular when its determinant is 0:
\( \displaystyle \begin{vmatrix}2 & k\\ k^{2} & 4\end{vmatrix} = 8 - k \cdot k^{2} \)
\(8 - k^{3} = 0\)
\(k^{3} = 8\)
The only real cube root of 8 is 2, so k = 2.
Why the other options are incorrect
Option (2) −2 — Incorrect. With k = −2 the determinant is \(8 - (-8) = 16 \ne 0\).
Option (3) 2 or −2 — Incorrect. This would come from \(k^{2} = 4\); the off-diagonal product is \(k \cdot k^{2} = k^{3}\), and k = −2 fails, as shown for Option (2).
Option (4) 4 — Incorrect. With k = 4 the determinant is \(8 - 64 = -56 \ne 0\).