1. **State the problem:** Solve the system of equations using the inverse matrix method: $$\begin{cases} x + 2y + 3z = 14 \\ 2x - y + z = 5 \\ 3x + y + 2z = 13 \end{cases}$$ 2. **Write the system in matrix form:** $$AX = B$$ where $$A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 14 \\ 5 \\ 13 \end{bmatrix}$$ 3. **Find the inverse of matrix $A$, denoted $A^{-1}$:** Calculate the determinant $\det(A)$: $$\det(A) = 1((-1)(2) - 1(1)) - 2(2(2) - 1(3)) + 3(2(1) - (-1)(3))$$ $$= 1(-2 - 1) - 2(4 - 3) + 3(2 + 3) = 1(-3) - 2(1) + 3(5) = -3 - 2 + 15 = 10$$ Since $\det(A) \neq 0$, $A$ is invertible. 4. **Calculate the matrix of minors, cofactors, and adjugate:** Minors: $$M = \begin{bmatrix} (-1)(2) - 1(1) & 2(2) - 1(3) & 2(1) - (-1)(3) \\ 2(2) - 3(1) & 1(2) - 3(3) & 1(1) - 2(3) \\ 2(-1) - 1(1) & 1(3) - 2(2) & 1(-1) - 2(2) \end{bmatrix} = \begin{bmatrix} -3 & 1 & 5 \\ 1 & -7 & -5 \\ -3 & -1 & -3 \end{bmatrix}$$ Cofactors (apply signs): $$C = \begin{bmatrix} -3 & -1 & 5 \\ -1 & -7 & 5 \\ -3 & 1 & -3 \end{bmatrix}$$ Adjugate (transpose of cofactors): $$\text{adj}(A) = C^T = \begin{bmatrix} -3 & -1 & -3 \\ -1 & -7 & 1 \\ 5 & 5 & -3 \end{bmatrix}$$ 5. **Calculate inverse:** $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{10} \begin{bmatrix} -3 & -1 & -3 \\ -1 & -7 & 1 \\ 5 & 5 & -3 \end{bmatrix}$$ 6. **Find $X$ by multiplying $A^{-1}$ and $B$:** $$X = A^{-1} B = \frac{1}{10} \begin{bmatrix} -3 & -1 & -3 \\ -1 & -7 & 1 \\ 5 & 5 & -3 \end{bmatrix} \begin{bmatrix} 14 \\ 5 \\ 13 \end{bmatrix}$$ Calculate each component: $$x = \frac{1}{10}(-3 \times 14 - 1 \times 5 - 3 \times 13) = \frac{1}{10}(-42 - 5 - 39) = \frac{1}{10}(-86) = -8.6$$ $$y = \frac{1}{10}(-1 \times 14 - 7 \times 5 + 1 \times 13) = \frac{1}{10}(-14 - 35 + 13) = \frac{1}{10}(-36) = -3.6$$ $$z = \frac{1}{10}(5 \times 14 + 5 \times 5 - 3 \times 13) = \frac{1}{10}(70 + 25 - 39) = \frac{1}{10}(56) = 5.6$$ 7. **Final answer:** $$X = -8.6, \quad Y = -3.6, \quad Z = 5.6$$