The system of equations
\(x + 2y - z = 0\)
\(3x + ky + z = 0\)
\(2x + y + 2z = 0\)
has a solution other than \(x = 0,\ y = 0,\ z = 0\). Then the value of k is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The system AX = O always has the solution x = y = z = 0. If \(|A| \ne 0\), this is its only solution. So a non-zero solution needs
\( \displaystyle |A| = \begin{vmatrix}1 & 2 & -1\\ 3 & k & 1\\ 2 & 1 & 2\end{vmatrix} = 0 \)
Expanding along the first row:
\( |A| = 1(2k - 1) - 2(6 - 2) + (-1)(3 - 2k) \)
\( = 2k - 1 - 8 - 3 + 2k \)
\( = 4k - 12 \)
\( 4k - 12 = 0 \)
\( k = 3 \)
Check: for k = 3, the values x = 5, y = −4, z = −3 satisfy all three equations: \(5 - 8 + 3 = 0\), \(15 - 12 - 3 = 0\) and \(10 - 4 - 6 = 0\).