1. **State the problem:** Find the inverse of matrix $A = \begin{pmatrix}-2 & 1 & 3 \\ 5 & 0 & 2 \\ 1 & 6 & -3\end{pmatrix}$. 2. **Formula and rules:** The inverse of a matrix $A$, denoted $A^{-1}$, satisfies $A \cdot A^{-1} = I$, where $I$ is the identity matrix. To find $A^{-1}$, we use the formula: $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ where $\det(A)$ is the determinant of $A$ and $\mathrm{adj}(A)$ is the adjugate matrix (transpose of the cofactor matrix). 3. **Calculate the determinant $\det(A)$:** $$\det(A) = -2 \begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} - 1 \begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} + 3 \begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} = 0 \times (-3) - 2 \times 6 = -12$$ $$\begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} = 5 \times (-3) - 2 \times 1 = -15 - 2 = -17$$ $$\begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix} = 5 \times 6 - 0 \times 1 = 30$$ Substitute back: $$\det(A) = -2(-12) - 1(-17) + 3(30) = 24 + 17 + 90 = 131$$ 4. **Find the cofactor matrix $C$:** Calculate each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor. - $C_{11} = (+1) \times \begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} = -12$ - $C_{12} = (-1) \times \begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} = -(-17) = 17$ - $C_{13} = (+1) \times \begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix} = 30$ - $C_{21} = (-1) \times \begin{vmatrix}1 & 3 \\ 6 & -3\end{vmatrix} = - (1 \times -3 - 3 \times 6) = -(-3 - 18) = 21$ - $C_{22} = (+1) \times \begin{vmatrix}-2 & 3 \\ 1 & -3\end{vmatrix} = (-2)(-3) - 3(1) = 6 - 3 = 3$ - $C_{23} = (-1) \times \begin{vmatrix}-2 & 1 \\ 1 & 6\end{vmatrix} = -((-2)(6) - 1(1)) = -(-12 - 1) = 13$ - $C_{31} = (+1) \times \begin{vmatrix}1 & 3 \\ 0 & 2\end{vmatrix} = 1 \times 2 - 3 \times 0 = 2$ - $C_{32} = (-1) \times \begin{vmatrix}-2 & 3 \\ 5 & 2\end{vmatrix} = -((-2)(2) - 3(5)) = -(-4 - 15) = 19$ - $C_{33} = (+1) \times \begin{vmatrix}-2 & 1 \\ 5 & 0\end{vmatrix} = (-2)(0) - 1(5) = -5$ So, $$C = \begin{pmatrix}-12 & 17 & 30 \\ 21 & 3 & 13 \\ 2 & 19 & -5\end{pmatrix}$$ 5. **Find the adjugate matrix $\mathrm{adj}(A)$:** Transpose the cofactor matrix: $$\mathrm{adj}(A) = C^T = \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}$$ 6. **Calculate the inverse:** $$A^{-1} = \frac{1}{131} \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}$$ **Final answer:** $$\boxed{A^{-1} = \frac{1}{131} \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}}$$ This inverse matrix satisfies $A \cdot A^{-1} = I$.