Let $S$ be the set of all $\lambda \in \mathbf{R}$ for which the system of linear equations
$2 x-y+2 z=2 \quad x-2 y+\lambda z=-4$
$x+\lambda y+z=4$
has no solution. Then the set $S$
Correct Option: , 4
$\Delta=\left|\begin{array}{ccc}2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1\end{array}\right|=-(\lambda-1)(2 \lambda+1)$
$\Delta_{1}=\left|\begin{array}{ccc}2 & -1 & 2 \\ -4 & -2 & \lambda \\ 4 & \lambda & 1\end{array}\right|=-2\left(\lambda^{2}+6 \lambda-4\right)$
For no solution $\Delta=0$ and at least one of $\Delta_{1}, \Delta_{2}$ and $\Delta_{3}$ is non-zero.
$\therefore \Delta=0 \Rightarrow \lambda=1,-\frac{1}{2}$ and $\Delta_{1} \neq 0$
Hence, $S=\left\{1,-\frac{1}{2}\right\}$
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