Question

The system of equation $\lambda x+y+z=0,-x+\lambda y+z=0,-x-y+\lambda z=0$ will have a non-zero solution if real values of $\lambda$ are given by ................

Answer 1

Given system $\lambda x+y+z=0,-x+\lambda y+z=0$ and $-x-y+\lambda z=0$
will have non-zero solution if
$\left|\begin{array}{ccc}\lambda & 1& 1\\ -1& \lambda & 1\\ -1& -1& \lambda \end{array}\right|=0\Rightarrow \lambda ({\lambda }^2+1)-1(-\lambda +1)+1(1+\lambda )=0\Rightarrow {\lambda }^3+\lambda +\lambda -1+1+\lambda =0\Rightarrow {\lambda }^3+3\lambda =0\Rightarrow \lambda ({\lambda }^2+3)=0\Rightarrow \lambda =0$