1. **Problem Statement:** Solve the system of linear equations using the inverse matrix method: $$\begin{cases} 3x + 2y - z = 5 \\ 2x - 2y + 4z = -2 \\ -x + 0y + 5z = 17 \end{cases}$$ 2. **Create the augmented matrix:** The coefficient matrix $A$ and constant vector $\mathbf{b}$ are: $$A = \begin{bmatrix} 3 & 2 & -1 \\ 2 & -2 & 4 \\ -1 & 0 & 5 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 5 \\ -2 \\ 17 \end{bmatrix}$$ The augmented matrix is: $$\left[ \begin{array}{ccc|c} 3 & 2 & -1 & 5 \\ 2 & -2 & 4 & -2 \\ -1 & 0 & 5 & 17 \end{array} \right]$$ 3. **Inverse matrix method formula:** To find $\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, use: $$\mathbf{x} = A^{-1} \mathbf{b}$$ where $A^{-1}$ is the inverse of matrix $A$. 4. **Calculate the inverse of $A$:** First, find the determinant $\det(A)$: $$\det(A) = 3 \begin{vmatrix} -2 & 4 \\ 0 & 5 \end{vmatrix} - 2 \begin{vmatrix} 2 & 4 \\ -1 & 5 \end{vmatrix} + (-1) \begin{vmatrix} 2 & -2 \\ -1 & 0 \end{vmatrix}$$ Calculate minors: $$= 3((-2)(5) - 0(4)) - 2(2 \cdot 5 - (-1) \cdot 4) - 1(2 \cdot 0 - (-1)(-2))$$ $$= 3(-10) - 2(10 + 4) - 1(0 - 2) = -30 - 2(14) - 1(-2) = -30 - 28 + 2 = -56$$ Since $\det(A) \neq 0$, $A$ is invertible. 5. **Find the adjugate matrix $\text{adj}(A)$:** Calculate cofactors and transpose: $$\text{adj}(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix}^T$$ Where cofactors $C_{ij} = (-1)^{i+j} M_{ij}$ and $M_{ij}$ is minor of element $a_{ij}$. Calculate minors: $M_{11} = \begin{vmatrix} -2 & 4 \\ 0 & 5 \end{vmatrix} = -10$ $M_{12} = \begin{vmatrix} 2 & 4 \\ -1 & 5 \end{vmatrix} = 2 \cdot 5 - (-1) \cdot 4 = 10 + 4 = 14$ $M_{13} = \begin{vmatrix} 2 & -2 \\ -1 & 0 \end{vmatrix} = 2 \cdot 0 - (-1)(-2) = 0 - 2 = -2$ $M_{21} = \begin{vmatrix} 2 & -1 \\ 0 & 5 \end{vmatrix} = 2 \cdot 5 - 0 \cdot (-1) = 10$ $M_{22} = \begin{vmatrix} 3 & -1 \\ -1 & 5 \end{vmatrix} = 3 \cdot 5 - (-1)(-1) = 15 - 1 = 14$ $M_{23} = \begin{vmatrix} 3 & 2 \\ -1 & 0 \end{vmatrix} = 3 \cdot 0 - (-1) \cdot 2 = 2$ $M_{31} = \begin{vmatrix} 2 & -1 \\ -2 & 4 \end{vmatrix} = 2 \cdot 4 - (-2)(-1) = 8 - 2 = 6$ $M_{32} = \begin{vmatrix} 3 & -1 \\ 2 & 4 \end{vmatrix} = 3 \cdot 4 - 2 \cdot (-1) = 12 + 2 = 14$ $M_{33} = \begin{vmatrix} 3 & 2 \\ 2 & -2 \end{vmatrix} = 3 \cdot (-2) - 2 \cdot 2 = -6 - 4 = -10$ Calculate cofactors: $C_{11} = (+1)(-10) = -10$ $C_{12} = (-1)(14) = -14$ $C_{13} = (+1)(-2) = -2$ $C_{21} = (-1)(10) = -10$ $C_{22} = (+1)(14) = 14$ $C_{23} = (-1)(2) = -2$ $C_{31} = (+1)(6) = 6$ $C_{32} = (-1)(14) = -14$ $C_{33} = (+1)(-10) = -10$ Transpose to get adjugate: $$\text{adj}(A) = \begin{bmatrix} -10 & -10 & 6 \\ -14 & 14 & -14 \\ -2 & -2 & -10 \end{bmatrix}$$ 6. **Calculate inverse matrix:** $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{-56} \begin{bmatrix} -10 & -10 & 6 \\ -14 & 14 & -14 \\ -2 & -2 & -10 \end{bmatrix}$$ 7. **Multiply $A^{-1}$ by $\mathbf{b}$:** $$\mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{-56} \begin{bmatrix} -10 & -10 & 6 \\ -14 & 14 & -14 \\ -2 & -2 & -10 \end{bmatrix} \begin{bmatrix} 5 \\ -2 \\ 17 \end{bmatrix}$$ Calculate the product inside the matrix: First row: $$-10 \cdot 5 + (-10) \cdot (-2) + 6 \cdot 17 = -50 + 20 + 102 = 72$$ Second row: $$-14 \cdot 5 + 14 \cdot (-2) + (-14) \cdot 17 = -70 - 28 - 238 = -336$$ Third row: $$-2 \cdot 5 + (-2) \cdot (-2) + (-10) \cdot 17 = -10 + 4 - 170 = -176$$ So: $$\mathbf{x} = \frac{1}{-56} \begin{bmatrix} 72 \\ -336 \\ -176 \end{bmatrix} = \begin{bmatrix} \frac{72}{-56} \\ \frac{-336}{-56} \\ \frac{-176}{-56} \end{bmatrix}$$ Simplify fractions: $$\frac{72}{-56} = \frac{\cancel{72}}{\cancel{56}} \cdot \frac{1}{-1} = \frac{9}{-7} = -\frac{9}{7}$$ $$\frac{-336}{-56} = \frac{\cancel{-336}}{\cancel{-56}} = 6$$ $$\frac{-176}{-56} = \frac{\cancel{-176}}{\cancel{-56}} = \frac{22}{7}$$ 8. **Final solution:** $$x = -\frac{9}{7}, \quad y = 6, \quad z = \frac{22}{7}$$ 9. **Advantages of inverse matrix method:** - Provides a direct formula for the solution. - Useful for theoretical analysis and computer algorithms. - Can be applied to any invertible square matrix. 10. **Conditions for applicability:** - The coefficient matrix must be square and invertible (non-zero determinant). 11. **Limitations and challenges:** - Computing inverse is computationally expensive for large matrices. - Numerical instability can occur with nearly singular matrices. - Not efficient for very large systems compared to other methods like LU decomposition.