Question

The system of equations

$\begin{array}{ccc}kx+y+z=1& & \\ x+ky+z=k& & \\ x+y+kz=k^2& & \end{array}$
have no solution,if k equals ?

• A. 0
• B. 1
• C. -1
• D. -2

Answer 1

The correct option is D -2

$\Delta =\left|\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right|$
$\Rightarrow \Delta =(k-1{)}^2(k+2)$
Taking $\Delta =0$
$\Rightarrow (k-1{)}^2(k+2)=0$
$\Rightarrow k=1,k=-2$
At these values of k, system can have either no solution or infinitely many solution.
For k=1, equations takes the form $x+y+z=1$
Hence, infinitely many solution for k=1.
For k=-2,
$D_1=\left|\begin{array}{ccc}1& 1& 1\\ -2& -2& 1\\ 4& 1& -2\end{array}\right|$
$D_1=9\ne 0$
So, $D=0$, at least one $D_1\ne 0$
Hence, system has no solution for k=-2