1. **Problem Statement:** Solve the system of simultaneous linear equations using the matrix inverse method: $$\begin{cases} 2x + 3y - z = 1 \\ -x + 2y + z = 8 \\ x - 3y - 2z = -13 \end{cases}$$ 2. **Matrix Form:** Write the system as $AX = B$ where $$A = \begin{bmatrix} 2 & 3 & -1 \\ -1 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 8 \\ -13 \end{bmatrix}$$ 3. **Find the inverse of matrix $A$:** Calculate $\det(A)$: $$\det(A) = 2 \begin{vmatrix} 2 & 1 \\ -3 & -2 \end{vmatrix} - 3 \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} -1 \begin{vmatrix} -1 & 2 \\ 1 & -3 \end{vmatrix}$$ Calculate minors: $$= 2(2 \times -2 - 1 \times -3) - 3(-1 \times -2 - 1 \times 1) - 1(-1 \times -3 - 2 \times 1)$$ $$= 2(-4 + 3) - 3(2 - 1) - 1(3 - 2) = 2(-1) - 3(1) - 1(1) = -2 - 3 - 1 = -6$$ 4. **Calculate the matrix of cofactors $C$:** $$C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}$$ Where each $C_{ij} = (-1)^{i+j} M_{ij}$ and $M_{ij}$ is the minor of element $a_{ij}$. Calculate minors: $M_{11} = \begin{vmatrix} 2 & 1 \\ -3 & -2 \end{vmatrix} = 2 \times -2 - 1 \times -3 = -4 + 3 = -1$ $M_{12} = \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} = -1 \times -2 - 1 \times 1 = 2 - 1 = 1$ $M_{13} = \begin{vmatrix} -1 & 2 \\ 1 & -3 \end{vmatrix} = -1 \times -3 - 2 \times 1 = 3 - 2 = 1$ $M_{21} = \begin{vmatrix} 3 & -1 \\ -3 & -2 \end{vmatrix} = 3 \times -2 - (-1) \times -3 = -6 - 3 = -9$ $M_{22} = \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = 2 \times -2 - (-1) \times 1 = -4 + 1 = -3$ $M_{23} = \begin{vmatrix} 2 & 3 \\ 1 & -3 \end{vmatrix} = 2 \times -3 - 3 \times 1 = -6 - 3 = -9$ $M_{31} = \begin{vmatrix} 3 & -1 \\ 2 & 1 \end{vmatrix} = 3 \times 1 - (-1) \times 2 = 3 + 2 = 5$ $M_{32} = \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} = 2 \times 1 - (-1) \times -1 = 2 - 1 = 1$ $M_{33} = \begin{vmatrix} 2 & 3 \\ -1 & 2 \end{vmatrix} = 2 \times 2 - 3 \times -1 = 4 + 3 = 7$ Apply signs: $$C = \begin{bmatrix} (-1)^{2}(-1) & (-1)^{3}(1) & (-1)^{4}(1) \\ (-1)^{3}(-9) & (-1)^{4}(-3) & (-1)^{5}(-9) \\ (-1)^{4}(5) & (-1)^{5}(1) & (-1)^{6}(7) \end{bmatrix} = \begin{bmatrix} -1 & -1 & 1 \\ 9 & -3 & 9 \\ 5 & -1 & 7 \end{bmatrix}$$ 5. **Transpose $C$ to get adjugate matrix $\text{adj}(A)$:** $$\text{adj}(A) = C^T = \begin{bmatrix} -1 & 9 & 5 \\ -1 & -3 & -1 \\ 1 & 9 & 7 \end{bmatrix}$$ 6. **Calculate inverse:** $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = -\frac{1}{6} \begin{bmatrix} -1 & 9 & 5 \\ -1 & -3 & -1 \\ 1 & 9 & 7 \end{bmatrix} = \begin{bmatrix} \frac{1}{6} & -\frac{3}{2} & -\frac{5}{6} \\ \frac{1}{6} & \frac{1}{2} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{3}{2} & -\frac{7}{6} \end{bmatrix}$$ 7. **Find solution vector $X = A^{-1}B$:** $$X = \begin{bmatrix} \frac{1}{6} & -\frac{3}{2} & -\frac{5}{6} \\ \frac{1}{6} & \frac{1}{2} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{3}{2} & -\frac{7}{6} \end{bmatrix} \begin{bmatrix} 1 \\ 8 \\ -13 \end{bmatrix}$$ Calculate each component: $x = \frac{1}{6} \times 1 - \frac{3}{2} \times 8 - \frac{5}{6} \times (-13) = \frac{1}{6} - 12 + \frac{65}{6} = \frac{1 + 65}{6} - 12 = \frac{66}{6} - 12 = 11 - 12 = -1$ $y = \frac{1}{6} \times 1 + \frac{1}{2} \times 8 + \frac{1}{6} \times (-13) = \frac{1}{6} + 4 - \frac{13}{6} = 4 + \frac{1 - 13}{6} = 4 - 2 = 2$ $z = -\frac{1}{6} \times 1 - \frac{3}{2} \times 8 - \frac{7}{6} \times (-13) = -\frac{1}{6} - 12 + \frac{91}{6} = -12 + \frac{-1 + 91}{6} = -12 + \frac{90}{6} = -12 + 15 = 3$ 8. **Final solution:** $$\boxed{x = -1, \quad y = 2, \quad z = 3}$$