Question

If the system of equations $x-ky-z=0,kx-y-z=0,x+y-z=0$ has a non-zero solution, then the possible values of $k$ are

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

Answer 1

The correct option is D

$-1,1$

Explanation for the correct option:

Find the possible values of $k$.

As the equations $x-ky-z=0,kx-y-z=0,x+y-z=0$ has a non-zero solution, so $D=0$.

$$\begin{gathered} \left|\begin{array}{ccc}1& -k& -1\\ k& -1& -1\\ 1& 1& -1\end{array}\right|=0 \\ \Rightarrow 1\left(1+1\right)+k\left(-k+1\right)-1\left(k+1\right)=0 \\ \Rightarrow 2-k^2+k-k-1=0 \\ \Rightarrow -k^2+1=0 \\ \Rightarrow k^2=1 \\ \Rightarrow k=\pm 1 \end{gathered}$$

Hence, the correct option is D.