Obtain the inverse of the following matrix, using the adjoint method. 4 7 2 3 6 1 2 5 3

1. Compute the determinant of the matrix \mathbf{A} .

\det(\mathbf{A}) = 4(6 \cdot 3 - 1 \cdot 5) - 7(3 \cdot 3 - 1 \cdot 2) + 2(3 \cdot 5 - 6 \cdot 2)

\det(\mathbf{A}) = 4(18 - 5) - 7(9 - 2) + 2(15 - 12)

\det(\mathbf{A}) = 4 \cdot 13 - 7 \cdot 7 + 2 \cdot 3

\det(\mathbf{A}) = 52 - 49 + 6 = 9

2. Compute the matrix of minors for \mathbf{A} .

\text{Minor}_{11} = \begin{vmatrix} 6 & 1 \\ 5 & 3 \end{vmatrix} = (6 \cdot 3 - 1 \cdot 5) = 18 - 5 = 13 \\\text{Minor}_{12} = \begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} = (3 \cdot 3 - 1 \cdot 2) = 9 - 2 = 7 \\\text{Minor}_{13} = \begin{vmatrix} 3 & 6 \\ 2 & 5 \end{vmatrix} = (3 \cdot 5 - 6 \cdot 2) = 15 - 12 = 3 \\\text{Minor}_{21} = \begin{vmatrix} 7 & 2 \\ 5 & 3 \end{vmatrix} = (7 \cdot 3 - 2 \cdot 5) = 21 - 10 = 11 \\\text{Minor}_{22} = \begin{vmatrix} 4 & 2 \\ 2 & 3 \end{vmatrix} = (4 \cdot 3 - 2 \cdot 2) = 12 - 4 = 8 \\\text{Minor}_{23} = \begin{vmatrix} 4 & 7 \\ 2 & 5 \end{vmatrix} = (4 \cdot 5 - 7 \cdot 2) = 20 - 14 = 6 \\\text{Minor}_{31} = \begin{vmatrix} 7 & 2 \\ 6 & 1 \end{vmatrix} = (7 \cdot 1 - 2 \cdot 6) = 7 - 12 = -5 \\\text{Minor}_{32} = \begin{vmatrix} 4 & 2 \\ 3 & 1 \end{vmatrix} = (4 \cdot 1 - 2 \cdot 3) = 4 - 6 = -2 \\\text{Minor}_{33} = \begin{vmatrix} 4 & 7 \\ 3 & 6 \end{vmatrix} = (4 \cdot 6 - 7 \cdot 3) = 24 - 21 = 3

3. Compute the matrix of cofactors for \mathbf{A} :

\text{Cofactor}_{ij} = (-1)^{i+j} \cdot \text{Minor}_{ij}

\mathbf{Cofactors}(A) = \begin{pmatrix}13 & -7 & 3 \\-11 & 8 & -6 \\-5 & 2 & 3\end{pmatrix}

4. Compute the adjugate (transpose of the cofactor matrix) for \mathbf{A} :

\mathbf{Adj}(A) = \begin{pmatrix}13 & -11 & -5 \\-7 & 8 & 2 \\3 & -6 & 3\end{pmatrix}

5. Compute the inverse of \mathbf{A} by multiplying the adjugate by the reciprocal of the determinant:

\mathbf{A}^{-1} = \frac{1}{9} \begin{pmatrix}13 & -11 & -5 \\-7 & 8 & 2 \\3 & -6 & 3\end{pmatrix}