Given

$2x-y +2z=2$

$x-2y +z= -4$

$x+y +\lambda z = 4 $

then the value of $\lambda $ such that the given system of equations has no solution, is

1

The correct answer is option (2) : 1

The given system of equations will have no solution, if D = 0 and at least one of $D_1, D_2, D_3 $ is non zero,

where

$D=\begin{vmatrix}2 & -1 & 2\\1 &-2  & 1\\1 & 1 & \lambda \end{vmatrix}, D_1= \begin{vmatrix}2 & -1 & 2\\-4 &-2  & 1\\4 & 1 & \lambda \end{vmatrix}, D_2= \begin{vmatrix}2 & 2 & 2\\1 &-4  & 1\\1 & 4 & \lambda \end{vmatrix}$

and, $D_3=\begin{vmatrix}2 & -1 & 2\\1 &-2  & -4\\1 & 1 & 4\end{vmatrix}$

Now,

$D=0$

$⇒\begin{vmatrix}2 & -1 & 2\\1 &-2  & 1\\1 & 1 & \lambda \end{vmatrix}=0$

$⇒2(-2\lambda-1) + ( \lambda - 1) + 2( 1+ 2) = 0 $

$⇒-3\lambda + 3= 0 $

$⇒\lambda = 1 $

For this value of $\lambda , $ we have

$D_1= \begin{vmatrix}2 & -1 & 2\\-4 &-2  & 1\\4 & 1 & 1\end{vmatrix}=6 -8 + 8 ≠ 0 $

Hence, the system has no solution for $\lambda = 1 $