If the system of equations $kx+y+z=0, x+ky-z=0, x-y+z=0$ has a non-zero solution, then the possible values of $k$ are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -1, 3 **
Coefficient matrix: \(\begin{pmatrix}k & 1 & 1\\[4pt]1 & k & -1\\[4pt]1 & -1 & 1\end{pmatrix}\)
Non-trivial solution ⇒ determinant = 0.
\(\det = k\begin{vmatrix}k & -1\\[4pt]-1 & 1\end{vmatrix}-1\begin{vmatrix}1 & -1\\[4pt]1 & 1\end{vmatrix}+1\begin{vmatrix}1 & k\\[4pt]1 & -1\end{vmatrix}\)
\(=k(k-1)-1(2)+1(-1-k)=k^{2}-k-2-1-k=k^{2}-2k-3\)
Set \(k^{2}-2k-3=0\Rightarrow (k-3)(k+1)=0\)
Hence \(k=3\) or \(k=-1\).