Question

If the system of equations $x-ky-z=0,kx-y-z=0$ and $x+y-z=0$
has a non-trivial solution, then $k$ can be

• A. $0,1$
• B. $1,-1$
• C. $-1,2$
• D. $2,-2$

Answer 1

The correct option is B $1,-1$

The given system of equation has a non-trivial solution if $\Delta =0$
For the given system of equations, we have
$\Delta =\left|\begin{array}{ccc}1& -k& -1\\ k& -1& -1\\ 1& 1& -1\end{array}\right|$
$C_1\to C_1+C_3$
Now, $\Delta =\left|\begin{array}{ccc}0& -k& -1\\ k-1& -1& -1\\ 0& 1& -1\end{array}\right|=0$
On solving, we get
$\Rightarrow (k-1)(k+1)=0\Rightarrow k=-1,1$