Mathematics - Determinant Question with Solution | TestHub

$kx+y+z=1$

$x+ky+z=k$

$x+y+zk=k^2$

$\Delta =\left|\begin{array}{ccc}K& 1& 1\\ 1& K& 1\\ 1& 1& K\end{array}\right|=K\left(K^2-1\right)-1\left(K-1\right)+1\left(1-K\right)$

$=K^3-K-K+1+1-K$

$=K^3-3 K+2$

$=(K-1{)}^2\left( K+2\right)$

For $K=1$

$\Delta ={\Delta }_1={\Delta }_2={\Delta }_3=0$

But for $K=-2,$ at least one out of ${\Delta }_1,{\Delta }_2,{\Delta }_3$ are not zero

Hence for no solution, $K=-2$