For the system of linear equations :
$x-2 y=1, x-y+k z=-2, k y+4 z=6, k \in R$
consider the following statements :
(A) The system has unique solution if $\mathrm{k} \neq 2$, $\mathrm{k} \neq-2$
(B) The system has unique solution if $\mathrm{k}=-2$.
(C) The system has unique solution if $\mathrm{k}=2$.
(D) The system has no-solution if $\mathrm{k}=2$.
(E) The system has infinite number of solutions if $\mathrm{k} \neq-2$.
Which of the following statements are correct?
Correct Option: , 4
$D=\left|\begin{array}{ccc}1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4\end{array}\right|=4-k^{2}$
so, $\mathrm{A}$ is correct and $\mathrm{B}, \mathrm{C}, \mathrm{E}$ are incorrect.
If $\mathrm{k}=2$
$D_{1}=\left|\begin{array}{ccc}1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4\end{array}\right|=-48 \neq 0$
So no solution
D is correct.
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