1. **Problem Statement:** 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 $A^{-1}$, exists only if $\det(A) \neq 0$. It satisfies the equation $$A \cdot A^{-1} = I,$$ where $I$ is the identity matrix. 3. **Step 1: Calculate the determinant of $A$.** $$\det(A) = -2 \begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} - 1 \begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} + 3 \begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} = (0)(-3) - (2)(6) = -12$$ $$\begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} = (5)(-3) - (2)(1) = -15 - 2 = -17$$ $$\begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix} = (5)(6) - (0)(1) = 30$$ So, $$\det(A) = -2(-12) - 1(-17) + 3(30) = 24 + 17 + 90 = 131$$ 4. **Step 2: Calculate the matrix of cofactors.** Calculate each cofactor $C_{ij} = (-1)^{i+j} M_{ij}$ where $M_{ij}$ is the minor. - $C_{11} = (+1) \times \begin{vmatrix}0 & 2 \\ 6 & -3\end{vmatrix} = -12$ - $C_{12} = (-1) \times \begin{vmatrix}5 & 2 \\ 1 & -3\end{vmatrix} = 17$ - $C_{13} = (+1) \times \begin{vmatrix}5 & 0 \\ 1 & 6\end{vmatrix} = 30$ - $C_{21} = (-1) \times \begin{vmatrix}1 & 3 \\ 6 & -3\end{vmatrix} = - (1 \times -3 - 3 \times 6) = -(-3 - 18) = 21$ - $C_{22} = (+1) \times \begin{vmatrix}-2 & 3 \\ 1 & -3\end{vmatrix} = (-2)(-3) - (3)(1) = 6 - 3 = 3$ - $C_{23} = (-1) \times \begin{vmatrix}-2 & 1 \\ 1 & 6\end{vmatrix} = -((-2)(6) - (1)(1)) = -(-12 - 1) = 13$ - $C_{31} = (+1) \times \begin{vmatrix}1 & 3 \\ 0 & 2\end{vmatrix} = (1)(2) - (3)(0) = 2$ - $C_{32} = (-1) \times \begin{vmatrix}-2 & 3 \\ 5 & 2\end{vmatrix} = -((-2)(2) - (3)(5)) = -(-4 - 15) = 19$ - $C_{33} = (+1) \times \begin{vmatrix}-2 & 1 \\ 5 & 0\end{vmatrix} = (-2)(0) - (1)(5) = -5$ 5. **Step 3: Form the cofactor matrix and transpose it (adjugate matrix).** $$\text{Cofactor matrix} = \begin{pmatrix}-12 & 17 & 30 \\ 21 & 3 & 13 \\ 2 & 19 & -5\end{pmatrix}$$ Transpose: $$\text{Adj}(A) = \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}$$ 6. **Step 4: Calculate the inverse matrix.** $$A^{-1} = \frac{1}{\det(A)} \text{Adj}(A) = \frac{1}{131} \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}$$ 7. **Final answer:** $$\boxed{A^{-1} = \frac{1}{131} \begin{pmatrix}-12 & 21 & 2 \\ 17 & 3 & 19 \\ 30 & 13 & -5\end{pmatrix}}$$ This inverse satisfies $A \cdot A^{-1} = I$, the identity matrix. **Note:** The statement $A \cdot A^{-1} = 2$ in the user message is incorrect; it should be the identity matrix $I$.