For the system of linear equations:
$x-2 y=1, x-y+k z=-2, k y+4 z=6, k \in \mathbf{R}$
consider the following statements:
(A) The system has unique solution if $k \neq 2, k \neq-2$.
(B) The system has unique solution if $k=-2$.
(C) The system has unique solution if $k=2$.
(D) The system has no-solution if $k=2$.
(E) The system has infinite number of solutions if $k \neq-2$.
Which of the following statements are correct?
Correct Option: , 3
$x-2 y+0 . z=1$
$x-y+k z=-2$
$0 \cdot x+k y+4 z=6$
$\Delta=\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{array}\right|=4-k^{2}$
For unique solution $4-\mathrm{k}^{2} \neq 0$
$\mathrm{k} \neq \pm 2$
For $\mathrm{k}=2$
$x-2 y+0 . z=1$
$x-y+2 z=-2$
$0 \cdot x+2 y+4 z=6$
$\Delta \mathrm{x}=\left|\begin{array}{ccc}1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4\end{array}\right|=(-8)+2[-20]$
$\Delta x=-48 \neq 0$
For $\mathrm{k}=2 \quad \Delta \mathrm{x} \neq 0$
For $\mathrm{K}=2$; The system has no solution