Q.The system of linear equations $x + y + z = 2$, $2x + y - z = 3$, $3x + 2y + k = 4$ has unique solution, if $k$ is not equal to :

A linear system $AX = B$ has a unique solution iff $|A| \neq 0$. Here $|A| = -k$, so uniqueness fails only when $k = 0$; hence a unique solution exists whenever $k \neq 0$.

Step 1 — Write the coefficient matrix of $x + y + z = 2,\; 2x + y - z = 3,\; 3x + 2y + kz = 4$:

$A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{pmatrix}.$

Step 2 — Expand $|A|$ along the first row.

$|A| = 1\big(1\cdot k - (-1)\cdot 2\big) - 1\big(2\cdot k - (-1)\cdot 3\big) + 1\big(2\cdot 2 - 1\cdot 3\big)$ …