Q.Solve the system $2x+3y=7,\ x-y=1$ using the matrix method.

A square system has a unique solution iff its coefficient determinant is non-zero; here that determinant is $(k-1)^2(k+2)$.

Cramer's-rule / determinant condition for a unique solution is a CBSE/NCERT Class 12 determinants and linear-equations topic.

The coefficient matrix is $\begin{bmatrix} k&1&1\\1&k&1\\1&1&k \end{bmatrix}$. Its determinant is

$\Delta = k(k^2-1) - 1(k-1) + 1(1-k) = k^3 - 3k + 2 = (k-1)^2(k+2).$

A unique solution exists precisely when $\Delta \neq 0$:

$(k-1)^2(k+2) \neq 0 \Rightarrow k \neq 1 \text{ and } k \neq -2.$

Eliminating $x$ from the system gives $y + 2z = 1$ and $y + 3z = 1$, so $z = 0$ and $y = 1$.

The system is

$x + y + z = 3 \quad (1), \qquad x + 2y + 3z = 4 \quad (2), \qquad x + 4y + 9z = 6 \quad (3).$

Subtract (1) from (2):

$(2) - (1): \; y + 2z = 1. \quad (4)$

Subtract (2) from (3):

$(3) - (2): \; 2y + 6z = 2 \;\Rightarrow\; y + 3z = 1. \quad (5)$

Subtract (4) from (5):

$(5) - (4): \; z = 0.$

Substitute $z = 0$ into (4): $y = 1$.