The system of equations $k x+y+z=1$, $x+k y+z=k$ and $x+y+z k=k^{2}$ has no solution if $\mathrm{k}$ is equal to :
Correct Option: , 4
$\mathrm{kx}+\mathrm{y}+\mathrm{z}=1$
$\mathrm{x}+\mathrm{ky}+\mathrm{z}=\mathrm{k}$
$\mathrm{x}+\mathrm{y}+\mathrm{zk}=\mathrm{k}^{2}$
$\Delta=\left|\begin{array}{ccc}\mathrm{K} & 1 & 1 \\ 1 & \mathrm{~K} & 1 \\ 1 & 1 & \mathrm{~K}\end{array}\right|=\mathrm{K}\left(\mathrm{K}^{2}-1\right)-1(\mathrm{~K}-1)+1(1-\mathrm{K})$
$=\mathrm{K}^{3}-\mathrm{K}-\mathrm{K}+1+1-\mathrm{K}$
$=\mathrm{K}^{3}-3 \mathrm{~K}+2$
$=(\mathrm{K}-1)^{2}(\mathrm{~K}+2)$
For $\mathrm{K}=1$
$\Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$
But for $\mathrm{K}=-2$, at least one out of $\Delta_{1}, \Delta_{2}, \Delta_{3}$ are not zero
Hence for no soln, $K=-2$