Step 1 (Formula / Governing Law): (1 Mark)

The inverse of a matrix $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$, provided $|A| \neq 0$. The system of linear equations $AX = B$ is solved by $X = A^{-1}B$.

Step 2 (Step-by-Step Working): (2 Marks)

Part 1: Finding $A^{-1}$

Calculate determinant $|A|$:

$$|A| = 1(0 - (-6)) - 2(0 - (-3)) - 3(4 - 0) = 1(6) - 2(3) - 3(4) = 6 - 6 - 12 = -12$$

Calculate the matrix of cofactors $C_{ij}$:

• $C_{11} = +(0 - (-6)) = 6, \quad C_{12} = -(0 - (-3)) = -3, \quad C_{13} = +(4 - 0) = 4$
• $C_{21} = -(0 - (-6)) = -6, \quad C_{22} = +(0 - (-3)) = 3, \quad C_{23} = -(2 - 2) = 0$
• $C_{31} = +(-6 - 0) = -6, \quad C_{32} = -(-3 - (-6)) = -3, \quad C_{33} = +(0 - 4) = -4$

$\text{adj}(A) = \begin{bmatrix} 6 & -6 & -6 \\ -3 & 3 & -3 \\ 4 & 0 & -4 \end{bmatrix}$

$$A^{-1} = -\frac{1}{12} \begin{bmatrix} 6 & -6 & -6 \\ -3 & 3 & -3 \\ 4 & 0 & -4 \end{bmatrix}$$

Part 2: Solving the System

The system is $AX = B$, where $B = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.

$$X = A^{-1}B = -\frac{1}{12} \begin{bmatrix} 6 & -6 & -6 \\ -3 & 3 & -3 \\ 4 & 0 & -4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$ $$X = -\frac{1}{12} \begin{bmatrix} 6 - 12 - 18 \\ -3 + 6 - 9 \\ 4 + 0 - 12 \end{bmatrix} = -\frac{1}{12} \begin{bmatrix} -24 \\ -6 \\ -8 \end{bmatrix}$$

Step 3 (Final Answer): (1 Mark)

$$x = \frac{24}{12} = 2, \quad y = \frac{6}{12} = \frac{1}{2}, \quad z