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

$ -1, 1$

The correct answer is option (4) : $ -1, 1$

The given system of equations has non-zero i.e.

non-trivial solution.

$∴\begin{bmatrix}1 & -k & -1\\k & -1 & -1\\1 & 1 & -1\end{bmatrix}=0$

$⇒\begin{bmatrix}1 & -k & -1\\k-1 & -1+k & 0\\0 & 1 +k & 0\end{bmatrix}=0$      $\begin{bmatrix}Applying \, R_2→R_2-R_1\\and\, R_3→R_3-R_1\end{bmatrix}$

$⇒ -(k^2-1) = 0 ⇒ k = \pm 1$