The value(s) of \(K\), for which the system of linear equations \(2x + y + z = 1, x+Ky-z=\displaystyle\frac{3}{2}\) and \(3y-5z = 9\) does not possess a unique solution is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(K=\displaystyle\frac{7}{5}\)
A system of three linear equations in three unknowns has a unique solution exactly when the determinant of the coefficient matrix is non-zero. So the solution is not unique when
\(\displaystyle \Delta = \begin{vmatrix}2&1&1\\1&K&-1\\0&3&-5\end{vmatrix} = 0\)
Expanding along the first row:
\(\Delta = 2(-5K + 3) - 1(-5 - 0) + 1(3 - 0)\)
\(= -10K + 6 + 5 + 3 = 14 - 10K\)
\(\displaystyle 14 - 10K = 0 \Rightarrow K = \frac{7}{5}\)
Why the other options are incorrect
Option (2) \(K=-\frac{7}{5}\) — Incorrect. Then \(\Delta = 14 + 14 = 28 \ne 0\), so the solution is unique.
Option (3) \(K \in R-\left\{\frac{7}{5}\right\}\) — Incorrect. For these values \(\Delta \ne 0\), which is exactly when the system does have a unique solution.
Option (4) \(K \in R-\left\{-\frac{7}{5}\right\}\) — Incorrect. This set contains, for example, K = 0, where \(\Delta = 14 \ne 0\) and the solution is unique.