• # question_answer If the solution of the linear equations$x-2y+z=0$; $2x-y+3z=0$ and$\lambda x+y-z=0$ trivial then the value of $\lambda$ is given by A) $\lambda =-\frac{4}{5}$                 B) $\lambda \ne -\frac{4}{5}$C) $\lambda =2$            D) $\lambda \ne 2$

For trivial solution,$\left| \begin{matrix} 1 & -2 & 1 \\ 2 & -1 & 3 \\ \lambda & 1 & -1 \\ \end{matrix} \right|\ne 0\Rightarrow -5\lambda -4\ne 0$or$\lambda \ne -\frac{4}{5}$