The integral value of k for which the system of linear equations \(kx + y + 2z = 0, ky=x-3z\) and \(2x + y + kz = 0\) has a non-zero solution is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
A homogeneous system has a non-zero solution only when the determinant of the coefficients is zero. Writing the second equation as \(x - ky - 3z = 0\):
\( \displaystyle \begin{vmatrix} k & 1 & 2 \\ 1 & -k & -3 \\ 2 & 1 & k \end{vmatrix} = 0 \)
Expanding along the first row:
\( k(-k^2 + 3) - 1(k + 6) + 2(1 + 2k) = 0 \)
\( -k^3 + 3k - k - 6 + 2 + 4k = 0 \)
\( k^3 - 6k + 4 = 0 \)
\( (k - 2)(k^2 + 2k - 2) = 0 \)
The quadratic factor gives \(k = -1 \pm \sqrt{3}\), which are not integers. So the integral value is
\( k = 2 \)