For some constant '\(k\)', if the system of linear equations
\(2x-y + 3z = 1\)
\(x-2y+z = 3\)
\(kx + y-z = 0\)
has a unique solution, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(k \ne -\displaystyle\frac{4}{5}\)
A system \(AX = B\) has a unique solution exactly when the coefficient determinant is non-zero.
\( \displaystyle |A| = \begin{vmatrix}2&-1&3\\1&-2&1\\k&1&-1\end{vmatrix} \)
Expanding along the first row:
\(|A| = 2(2 - 1) + 1(-1 - k) + 3(1 + 2k)\)
\(= 2 - 1 - k + 3 + 6k\)
\(= 4 + 5k\)
\(4 + 5k \ne 0\)
\( \displaystyle k \ne -\frac{4}{5} \)