1. **State the problem:** Find the inverse of matrix $$A=\begin{bmatrix}2 & 0 & 6 \\ 3 & 2 & 7 \\ 1 & 1 & 3\end{bmatrix}$$ 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 cofactor matrix). 3. **Calculate the determinant $$\det(A)$$:** $$\det(A) = 2 \times \begin{vmatrix}2 & 7 \\ 1 & 3\end{vmatrix} - 0 \times \begin{vmatrix}3 & 7 \\ 1 & 3\end{vmatrix} + 6 \times \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}2 & 7 \\ 1 & 3\end{vmatrix} = 2 \times 3 - 7 \times 1 = 6 - 7 = -1$$ $$\begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 3 \times 1 - 2 \times 1 = 3 - 2 = 1$$ So, $$\det(A) = 2 \times (-1) - 0 + 6 \times 1 = -2 + 6 = 4$$ 4. **Calculate the cofactor matrix:** Cofactor $$C_{ij} = (-1)^{i+j} M_{ij}$$ where $$M_{ij}$$ is the minor of element $$a_{ij}$$. Calculate each cofactor: - $$C_{11} = (+1) \times \begin{vmatrix}2 & 7 \\ 1 & 3\end{vmatrix} = -1$$ - $$C_{12} = (-1) \times \begin{vmatrix}3 & 7 \\ 1 & 3\end{vmatrix} = - (3 \times 3 - 7 \times 1) = -(9 - 7) = -2$$ - $$C_{13} = (+1) \times \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1$$ - $$C_{21} = (-1) \times \begin{vmatrix}0 & 6 \\ 1 & 3\end{vmatrix} = - (0 \times 3 - 6 \times 1) = - (0 - 6) = 6$$ - $$C_{22} = (+1) \times \begin{vmatrix}2 & 6 \\ 1 & 3\end{vmatrix} = 2 \times 3 - 6 \times 1 = 6 - 6 = 0$$ - $$C_{23} = (-1) \times \begin{vmatrix}2 & 0 \\ 1 & 1\end{vmatrix} = - (2 \times 1 - 0 \times 1) = -2$$ - $$C_{31} = (+1) \times \begin{vmatrix}0 & 6 \\ 2 & 7\end{vmatrix} = 0 \times 7 - 6 \times 2 = -12$$ - $$C_{32} = (-1) \times \begin{vmatrix}2 & 6 \\ 3 & 7\end{vmatrix} = - (2 \times 7 - 6 \times 3) = - (14 - 18) = 4$$ - $$C_{33} = (+1) \times \begin{vmatrix}2 & 0 \\ 3 & 2\end{vmatrix} = 2 \times 2 - 0 \times 3 = 4$$ 5. **Form the cofactor matrix:** $$C = \begin{bmatrix}-1 & -2 & 1 \\ 6 & 0 & -2 \\ -12 & 4 & 4\end{bmatrix}$$ 6. **Find the adjoint matrix $$\operatorname{adj}(A)$$ by transposing the cofactor matrix:** $$\operatorname{adj}(A) = C^T = \begin{bmatrix}-1 & 6 & -12 \\ -2 & 0 & 4 \\ 1 & -2 & 4\end{bmatrix}$$ 7. **Calculate the inverse:** $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1}{4} \begin{bmatrix}-1 & 6 & -12 \\ -2 & 0 & 4 \\ 1 & -2 & 4\end{bmatrix}$$ 8. **Simplify by dividing each element by 4:** $$A^{-1} = \begin{bmatrix}-\frac{1}{4} & \frac{6}{4} & -3 \\ -\frac{1}{2} & 0 & 1 \\ \frac{1}{4} & -\frac{1}{2} & 1\end{bmatrix}$$ This is the inverse matrix of $$A$$ using the adjoint method.