For some constant '$k$', if the system of linear equations
$2x-y + 3z = 1$
$x-2y+z = 3$
$kx + y-z = 0$
has a unique solution, then

$k≠-\frac{4}{5}$

The correct answer is Option (4) → $k≠-\frac{4}{5}$

$\text{Coefficient matrix}=\begin{pmatrix}2&-1&3\\[4pt]1&-2&1\\[4pt]k&1&-1\end{pmatrix}$

$\det=\begin{vmatrix}2&-1&3\\[4pt]1&-2&1\\[4pt]k&1&-1\end{vmatrix}$

$=2\bigl((-2)(-1)-1\cdot1\bigr)\;+\;1\bigl((-1)-k\bigr)\;+\;3\bigl(1+2k\bigr)$

$=2(1)\;+\;(-1-k)\;+\;(3+6k)$

$=4+5k$

$\text{For unique solution: }4+5k\neq0$

The required condition is $k\neq-\frac{4}{5}$.