Question

For the system of linear equations :
$x-2y=1,x-y+kz=-2,ky+4z=6,k\in R,$
consider the following statements :
$\text{(}A)$ The system has unique solution if $k\ne 2,k\ne -2.$
$\left(B\right)$ The system has unique solution if $k=-2.$
$\left(C\right)$ The system has unique solution if $k=2.$
$\left(D\right)$ The system has no-solution if $k=2.$
$\left(E\right)$ The system has infinite number of solutions if $k\ne -2.$

Which of the following statements are correct ?

• A. $\left(B\right)$ and $\left(E\right)$ only
• B. $\left(C\right)$ and $\left(D\right)$ only
• C. $\left(A\right)$ and $\left(D\right)$ only
• D. $\left(A\right)$ and $\left(E\right)$ only

Answer 1

The correct option is C $\left(A\right)$ and $\left(D\right)$ only

$x-2y+0\cdot z=1$
$x-y+kz=-2$
$0\cdot x+ky+4z=6$
$\Delta =\left|\begin{array}{ccc}1& -2& 0\\ 1& -1& k\\ 0& k& 4\end{array}\right|=4-k^2$
For unique solution,
$4-k^2\ne 0$
$\Rightarrow k\ne \pm 2$

For $k=2,$
$x-2y+0\cdot z=1$
$x-y+2z=-2$
$0\cdot x+2y+4z=6$
${\Delta }_x=\left|\begin{array}{ccc}1& -2& 0\\ -2& -1& 2\\ 6& 2& 4\end{array}\right|=(-8)+2(-20)$
$\Rightarrow {\Delta }_x=-48\ne 0$
For $k=2,{\Delta }_x\ne 0$
So, for $k=2,$ the system has no solution.