Q.Solve the following equations by the method of reduction:
$x + 3y = 2$, $3x + 5y = 4$

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

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

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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): …