The value of k for which the following system of equations does not possess a unique solution is:

$k x+3 y-z=1$

$x+2 y+z=2$

$-k x+y+2 z=-1$

$-\frac{7}{2}$

The correct answer is Option (1) → $-\frac{7}{2}$

The given system of equations,

$k x+3 y-z=1$

$x+2 y+z=2$

$-k x+y+2 z=-1$

Coefficient Matrix, $A=\begin{bmatrix}k&33&-1\\1&2&1\\-k&1&2\end{bmatrix}$

To ensure system doesn't have a solution, $|A|=0$

$∴\begin{bmatrix}k&33&-1\\1&2&1\\-k&1&2\end{bmatrix}=0$

$⇒3k-3(k+2)-(1+2k)=0$

$⇒-7-2k=0$

$⇒k=-\frac{7}{2}$