1. **Problem Statement:** Find the inverse of the matrix $$A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix}$$ using the adjoint method. 2. **Formula and Important Rules:** The inverse of a matrix $A$ is given by $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$$ where $\det(A)$ is the determinant of $A$ and $\operatorname{adj}(A)$ is the adjoint (transpose of the cofactor matrix). 3. **Step 1: Calculate the determinant $\det(A)$:** $$\det(A) = 1 \times \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} - 0 \times \begin{vmatrix} -1 & 2 \\ 1 & 2 \end{vmatrix} + 1 \times \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix}$$ Calculate minors: $$= 1 \times (2 \times 2 - 2 \times 1) + 0 + 1 \times (-1 \times 1 - 2 \times 1)$$ $$= 1 \times (4 - 2) + 0 + 1 \times (-1 - 2)$$ $$= 1 \times 2 + 0 - 3 = 2 - 3 = -1$$ 4. **Step 2: Find the cofactor matrix:** Calculate each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor of element $a_{ij}$. - $C_{11} = (+1) \times \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = 4 - 2 = 2$ - $C_{12} = (-1) \times \begin{vmatrix} -1 & 2 \\ 1 & 2 \end{vmatrix} = -((-1)(2) - 2(1)) = -( -2 - 2) = 4$ - $C_{13} = (+1) \times \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1)(1) - 2(1) = -1 - 2 = -3$ - $C_{21} = (-1) \times \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = - (0 \times 2 - 1 \times 1) = - (0 - 1) = 1$ - $C_{22} = (+1) \times \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 1 \times 2 - 1 \times 1 = 2 - 1 = 1$ - $C_{23} = (-1) \times \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = - (1 \times 1 - 0 \times 1) = - (1 - 0) = -1$ - $C_{31} = (+1) \times \begin{vmatrix} 0 & 1 \\ 2 & 2 \end{vmatrix} = 0 \times 2 - 1 \times 2 = 0 - 2 = -2$ - $C_{32} = (-1) \times \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = - (1 \times 2 - 1 \times (-1)) = - (2 + 1) = -3$ - $C_{33} = (+1) \times \begin{vmatrix} 1 & 0 \\ -1 & 2 \end{vmatrix} = 1 \times 2 - 0 \times (-1) = 2 - 0 = 2$ So the cofactor matrix is: $$\begin{bmatrix} 2 & 4 & -3 \\ 1 & 1 & -1 \\ -2 & -3 & 2 \end{bmatrix}$$ 5. **Step 3: Find the adjoint matrix $\operatorname{adj}(A)$:** Transpose the cofactor matrix: $$\operatorname{adj}(A) = \begin{bmatrix} 2 & 1 & -2 \\ 4 & 1 & -3 \\ -3 & -1 & 2 \end{bmatrix}$$ 6. **Step 4: Calculate the inverse matrix:** $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{-1} \begin{bmatrix} 2 & 1 & -2 \\ 4 & 1 & -3 \\ -3 & -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & -1 & 2 \\ -4 & -1 & 3 \\ 3 & 1 & -2 \end{bmatrix}$$ **Final answer:** $$\boxed{A^{-1} = \begin{bmatrix} -2 & -1 & 2 \\ -4 & -1 & 3 \\ 3 & 1 & -2 \end{bmatrix}}$$