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
A homogeneous system has a non-zero solution only when the determinant of its coefficients is 0:
\( \displaystyle \begin{vmatrix}k&1&1\\1&k&-1\\1&-1&1\end{vmatrix} = 0 \)
Expanding along the first row:
\(k(k \cdot 1 - (-1)(-1)) - 1(1 \cdot 1 - (-1) \cdot 1) + 1(1 \cdot (-1) - k \cdot 1) = 0\)
\(k(k - 1) - 2 + (-1 - k) = 0\)
\(k^2 - 2k - 3 = 0\)
\((k - 3)(k + 1) = 0\)
So \(k = 3\) or \(k = -1\).