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
Answer & explanation
Correct answer: option 4
The given system of equations has non-zero i.e. non-trivial solution.
$∴\begin{vmatrix}1&-k&-1\\k&-1&-1\\1&1&-1\end{vmatrix}=0$
$⇒\begin{vmatrix}1&-k&-1\\k-1&-1+k&0\\0&1+k&0\end{vmatrix}=0$ [Applying $R_2 → R_2-R_1$ and $R_3 →R_3-R_1$]
$⇒-(k^2-1)=0⇒ k=±1$