If the system of linear equations \(x+y+z=2\), \(2x+y-z=3\) and \(3x+2y+kz=4\) has a unique solution, then:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(k \ne 0\)
A system of three linear equations in three unknowns has a unique solution exactly when the coefficient determinant is non-zero.
\( \displaystyle \Delta = \begin{vmatrix}1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 2 & k\end{vmatrix} \)
Expanding along the first row:
\(\Delta = 1(k + 2) - 1(2k + 3) + 1(4 - 3)\)
\(= k + 2 - 2k - 3 + 1\)
\(= -k\)
So \(\Delta \ne 0\) exactly when \(k \ne 0\).
Why the other options are incorrect
Option (1) \(k=0\) — Incorrect. At k = 0, Δ = 0, so the system does not have a unique solution.
Option (2) \(-1 \lt k \lt 1\) — Incorrect. This range contains k = 0, where Δ = 0; it also leaves out values such as k = 2, which give a unique solution.
Option (4) \(-3 \lt k \lt 3\) — Incorrect. This range also contains k = 0, and it leaves out values such as k = 5, which give a unique solution.