The value of k for which the following system of equations does not possess a unique solution is:
$k x+3 y-z=1$
$x+2 y+z=2$
$-k x+y+2 z=-1$
Answer & explanation
Correct answer: option 1
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, $|A|=0$
$∴\begin{bmatrix}k&33&-1\\1&2&1\\-k&1&2\end{bmatrix}=0$
$⇒3k-3(k+2)-(1+2k)=0$
$⇒-7-2k=0$
$⇒k=-\frac{7}{2}$