If \( A=\begin{bmatrix}2 & 1 & -1\\0 & 3 & -2\\k & 1 & 0\end{bmatrix} \) is non-singular matrix, then the value (s) of \( k \) belongs to:
[Where \( R \) is the set of real numbers]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( R-\{-4\} \)
A is non-singular when \( |A|\ne 0 \). Expanding along the first row:
\( \displaystyle |A|=2\begin{vmatrix}3 & -2\\1 & 0\end{vmatrix}-1\begin{vmatrix}0 & -2\\k & 0\end{vmatrix}+(-1)\begin{vmatrix}0 & 3\\k & 1\end{vmatrix} \)
\( \displaystyle =2(0+2)-(0+2k)-(0-3k) \)
\( \displaystyle =4-2k+3k \)
\( \displaystyle =4+k \)
\( |A|\ne 0\ \Rightarrow\ k\ne -4 \)
So \( k\in R-\{-4\} \).
Why the other options are incorrect
Option (1) \( R-\{0\} \) — Incorrect. At k = 0, |A| = 4, which is not zero; A is singular only at k = −4.
Option (2) \( R-\{4\} \) — Incorrect. At k = 4, |A| = 8; this drops the sign, since 4 + k = 0 gives k = −4.
Option (4) \( R \) — Incorrect. At k = −4, |A| = 0, so not every real k works.