1. **State the problem:** Solve the system of equations using the inverse matrix method: $$\begin{cases} x - y + z = 8 \\ 2x + y - 3z = -15 \\ x + y + z = 4 \end{cases}$$ 2. **Write the system in matrix form:** $$AX = B$$ where $$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ -15 \\ 4 \end{bmatrix}$$ 3. **Find the inverse of matrix $A$, denoted $A^{-1}$:** Calculate the determinant of $A$: $$\det(A) = 1 \times \begin{vmatrix} 1 & -3 \\ 1 & 1 \end{vmatrix} - (-1) \times \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} + 1 \times \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix}$$ $$= 1(1 \times 1 - (-3) \times 1) + 1(2 \times 1 - (-3) \times 1) + 1(2 \times 1 - 1 \times 1)$$ $$= 1(1 + 3) + 1(2 + 3) + 1(2 - 1) = 4 + 5 + 1 = 10$$ Since $\det(A) \neq 0$, $A$ is invertible. 4. **Compute the adjugate matrix $\text{adj}(A)$ and then $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$:** Calculate cofactors and transpose to get adjugate: $$\text{adj}(A) = \begin{bmatrix} 4 & 5 & 1 \\ 2 & 0 & -2 \\ 3 & -7 & 3 \end{bmatrix}$$ Therefore, $$A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 5 & 1 \\ 2 & 0 & -2 \\ 3 & -7 & 3 \end{bmatrix}$$ 5. **Find $X$ by multiplying $A^{-1}$ and $B$:** $$X = A^{-1} B = \frac{1}{10} \begin{bmatrix} 4 & 5 & 1 \\ 2 & 0 & -2 \\ 3 & -7 & 3 \end{bmatrix} \begin{bmatrix} 8 \\ -15 \\ 4 \end{bmatrix}$$ Calculate each component: $$x = \frac{1}{10} (4 \times 8 + 5 \times (-15) + 1 \times 4) = \frac{1}{10} (32 - 75 + 4) = \frac{-39}{10} = -3.9$$ $$y = \frac{1}{10} (2 \times 8 + 0 \times (-15) + (-2) \times 4) = \frac{1}{10} (16 + 0 - 8) = \frac{8}{10} = 0.8$$ $$z = \frac{1}{10} (3 \times 8 + (-7) \times (-15) + 3 \times 4) = \frac{1}{10} (24 + 105 + 12) = \frac{141}{10} = 14.1$$ 6. **Final answer:** $$x = -3.9, \quad y = 0.8, \quad z = 14.1$$