1. **State the problem:** 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 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. **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: $$\begin{vmatrix}2 & 2 \\ 1 & 2\end{vmatrix} = (2)(2) - (2)(1) = 4 - 2 = 2$$ $$\begin{vmatrix}-1 & 2 \\ 1 & 1\end{vmatrix} = (-1)(1) - (2)(1) = -1 - 2 = -3$$ So, $$\det(A) = 1 \times 2 + 0 + 1 \times (-3) = 2 - 3 = -1$$ 4. **Find the cofactor matrix:** Calculate each cofactor $$C_{ij} = (-1)^{i+j} M_{ij}$$ where $$M_{ij}$$ is the minor of element $$a_{ij}$$. - $$C_{11} = (+1) \times \begin{vmatrix}2 & 2 \\ 1 & 2\end{vmatrix} = 2$$ - $$C_{12} = (-1) \times \begin{vmatrix}-1 & 2 \\ 1 & 2\end{vmatrix} = -((-1)(2) - (2)(1)) = -(-2 - 2) = 4$$ - $$C_{13} = (+1) \times \begin{vmatrix}-1 & 2 \\ 1 & 1\end{vmatrix} = -3$$ - $$C_{21} = (-1) \times \begin{vmatrix}0 & 1 \\ 1 & 2\end{vmatrix} = -((0)(2) - (1)(1)) = -(-1) = 1$$ - $$C_{22} = (+1) \times \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = (1)(2) - (1)(1) = 2 - 1 = 1$$ - $$C_{23} = (-1) \times \begin{vmatrix}1 & 0 \\ 1 & 1\end{vmatrix} = -((1)(1) - (0)(1)) = -1$$ - $$C_{31} = (+1) \times \begin{vmatrix}0 & 1 \\ 2 & 2\end{vmatrix} = (0)(2) - (1)(2) = -2$$ - $$C_{32} = (-1) \times \begin{vmatrix}1 & 1 \\ -1 & 2\end{vmatrix} = -((1)(2) - (1)(-1)) = -(2 + 1) = -3$$ - $$C_{33} = (+1) \times \begin{vmatrix}1 & 0 \\ -1 & 2\end{vmatrix} = (1)(2) - (0)(-1) = 2$$ 5. **Form the cofactor matrix:** $$C = \begin{bmatrix}2 & 4 & -3 \\ 1 & 1 & -1 \\ -2 & -3 & 2\end{bmatrix}$$ 6. **Find the adjoint matrix $$\operatorname{adj}(A)$$ by transposing the cofactor matrix:** $$\operatorname{adj}(A) = C^T = \begin{bmatrix}2 & 1 & -2 \\ 4 & 1 & -3 \\ -3 & -1 & 2\end{bmatrix}$$ 7. **Calculate the inverse:** $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{-1} \begin{bmatrix}2 & 1 & -2 \\ 4 & 1 & -3 \\ -3 & -1 & 2\end{bmatrix} = \begin{bmatrix}-2 & -1 & 2 \\ -4 & -1 & 3 \\ 3 & 1 & -2\end{bmatrix}$$ **Final answer:** $$A^{-1} = \begin{bmatrix}-2 & -1 & 2 \\ -4 & -1 & 3 \\ 3 & 1 & -2\end{bmatrix}$$