1. **Problem:** Find the inverse of matrix $$A = \begin{pmatrix}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{pmatrix}$$. 2. **Formula and rules:** The inverse of a matrix $$A$$, denoted $$A^{-1}$$, exists only if $$\det(A) \neq 0$$. The inverse satisfies $$AA^{-1} = I$$ where $$I$$ is the identity matrix. 3. **Step 1: Calculate the determinant of $$A$$.** $$\det(A) = 1 \times \begin{vmatrix}5 & 3 \\ 0 & 8\end{vmatrix} - 2 \times \begin{vmatrix}2 & 3 \\ 1 & 8\end{vmatrix} + 3 \times \begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}5 & 3 \\ 0 & 8\end{vmatrix} = 5 \times 8 - 0 \times 3 = 40$$ $$\begin{vmatrix}2 & 3 \\ 1 & 8\end{vmatrix} = 2 \times 8 - 1 \times 3 = 16 - 3 = 13$$ $$\begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix} = 2 \times 0 - 1 \times 5 = -5$$ So, $$\det(A) = 1 \times 40 - 2 \times 13 + 3 \times (-5) = 40 - 26 - 15 = -1$$ Since $$\det(A) = -1 \neq 0$$, the inverse exists. 4. **Step 2: Find the matrix of cofactors.** Calculate each cofactor $$C_{ij} = (-1)^{i+j} M_{ij}$$ where $$M_{ij}$$ is the minor. Cofactors: $$C_{11} = + \begin{vmatrix}5 & 3 \\ 0 & 8\end{vmatrix} = 40$$ $$C_{12} = - \begin{vmatrix}2 & 3 \\ 1 & 8\end{vmatrix} = -13$$ $$C_{13} = + \begin{vmatrix}2 & 5 \\ 1 & 0\end{vmatrix} = -5$$ $$C_{21} = - \begin{vmatrix}2 & 3 \\ 0 & 8\end{vmatrix} = -(2 \times 8 - 0) = -16$$ $$C_{22} = + \begin{vmatrix}1 & 3 \\ 1 & 8\end{vmatrix} = 1 \times 8 - 1 \times 3 = 5$$ $$C_{23} = - \begin{vmatrix}1 & 2 \\ 1 & 0\end{vmatrix} = -(1 \times 0 - 1 \times 2) = 2$$ $$C_{31} = + \begin{vmatrix}2 & 3 \\ 5 & 3\end{vmatrix} = 2 \times 3 - 5 \times 3 = 6 - 15 = -9$$ $$C_{32} = - \begin{vmatrix}1 & 3 \\ 2 & 3\end{vmatrix} = -(1 \times 3 - 2 \times 3) = -(3 - 6) = 3$$ $$C_{33} = + \begin{vmatrix}1 & 2 \\ 2 & 5\end{vmatrix} = 1 \times 5 - 2 \times 2 = 5 - 4 = 1$$ 5. **Step 3: Form the cofactor matrix and transpose it (adjugate matrix).** $$\text{Cofactor matrix} = \begin{pmatrix}40 & -13 & -5 \\ -16 & 5 & 2 \\ -9 & 3 & 1\end{pmatrix}$$ Transpose: $$\text{Adj}(A) = \begin{pmatrix}40 & -16 & -9 \\ -13 & 5 & 3 \\ -5 & 2 & 1\end{pmatrix}$$ 6. **Step 4: Calculate the inverse using $$A^{-1} = \frac{1}{\det(A)} \text{Adj}(A)$$** Since $$\det(A) = -1$$, $$A^{-1} = -1 \times \begin{pmatrix}40 & -16 & -9 \\ -13 & 5 & 3 \\ -5 & 2 & 1\end{pmatrix} = \begin{pmatrix}-40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1\end{pmatrix}$$ **Final answer:** $$A^{-1} = \begin{pmatrix}-40 & 16 & 9 \\ 13 & -5 & -3 \\ 5 & -2 & -1\end{pmatrix}$$