The value(s) of $K$, for which the system of linear equations $2x + y + z = 1, x + Ky - z = \frac{3}{2}$ and $3y - 5z = 9$ does not possess a unique solution is

A system of linear equations has a unique solution when the determinant of its coefficient matrix is non-zero. For the system to not possess a unique solution, the determinant must equal zero.

From the given equations:

$2x + y + z = 1$

$x + Ky - z = \dfrac{3}{2}$

$0 \cdot x + 3y - 5z = 9$

The coefficient matrix is:

$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & K & -1 \\ 0 & 3 & -5 \end{bmatrix}$

Note: The third equation has no $x$ term, so the first entry in the third row is $0$.

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Expanding along the first column (since it contains a zero, one term of computation is eliminated):

$\det(A) = 2\begin{vmatrix} K & -1 \\ 3 & -5 \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 3 & -5 \end{vmatrix} + 0\begin{vmatrix} 1 & 1 \\ K & -1 \end{vmatrix}$

Recall that for a $2 \times 2$ determinant: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$

Computing each $2 \times 2$ determinant:

$\begin{vmatrix} K & -1 \\ 3 & -5 \end{vmatrix} = (K)(-5) - (-1)(3) = -5K + 3$

$\begin{vmatrix} 1 & 1 \\ 3 & -5 \end{vmatrix} = (1)(-5) - (1)(3) = -8$

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Substituting back:

$\det(A) = 2(-5K + 3) - 1(-8) + 0$

$= -10K + 6 + 8$

$= -10K + 14$

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For no unique solution, $\det(A) = 0$:

$-10K + 14 = 0$

$K = \dfrac{14}{10}$

$K = \dfrac{7}{5}$

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Therefore, the system does not possess a unique solution when

$K = \dfrac{7}{5}$