1. **State the problem:** Find the inverse of the matrix $$A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 6 & 7 & 8\end{bmatrix}$$. 2. **Formula and rules:** The inverse of a matrix $$A$$ exists only if $$\det(A) \neq 0$$. The inverse is given by $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$, where $$\mathrm{adj}(A)$$ is the adjugate matrix. 3. **Calculate the determinant:** $$\det(A) = 1 \times \begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} - 2 \times \begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} + 3 \times \begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} = 4 \times 8 - 5 \times 7 = 32 - 35 = -3$$ $$\begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} = 0 \times 8 - 5 \times 6 = 0 - 30 = -30$$ $$\begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix} = 0 \times 7 - 4 \times 6 = 0 - 24 = -24$$ So, $$\det(A) = 1 \times (-3) - 2 \times (-30) + 3 \times (-24) = -3 + 60 - 72 = -15$$ 4. **Calculate the matrix of cofactors:** Each cofactor $$C_{ij} = (-1)^{i+j} M_{ij}$$ where $$M_{ij}$$ is the minor of element $$a_{ij}$$. Calculate all minors: - $$M_{11} = \begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} = -3$$ - $$M_{12} = \begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} = -30$$ - $$M_{13} = \begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix} = -24$$ - $$M_{21} = \begin{vmatrix}2 & 3 \\ 7 & 8\end{vmatrix} = 2 \times 8 - 3 \times 7 = 16 - 21 = -5$$ - $$M_{22} = \begin{vmatrix}1 & 3 \\ 6 & 8\end{vmatrix} = 1 \times 8 - 3 \times 6 = 8 - 18 = -10$$ - $$M_{23} = \begin{vmatrix}1 & 2 \\ 6 & 7\end{vmatrix} = 1 \times 7 - 2 \times 6 = 7 - 12 = -5$$ - $$M_{31} = \begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = 2 \times 5 - 3 \times 4 = 10 - 12 = -2$$ - $$M_{32} = \begin{vmatrix}1 & 3 \\ 0 & 5\end{vmatrix} = 1 \times 5 - 3 \times 0 = 5 - 0 = 5$$ - $$M_{33} = \begin{vmatrix}1 & 2 \\ 0 & 4\end{vmatrix} = 1 \times 4 - 2 \times 0 = 4 - 0 = 4$$ Apply signs for cofactors: $$C = \begin{bmatrix} (-1)^{2}(-3) & (-1)^{3}(-30) & (-1)^{4}(-24) \\ (-1)^{3}(-5) & (-1)^{4}(-10) & (-1)^{5}(-5) \\ (-1)^{4}(-2) & (-1)^{5}(5) & (-1)^{6}(4) \end{bmatrix} = \begin{bmatrix} -3 & 30 & -24 \\ 5 & -10 & 5 \\ -2 & -5 & 4 \end{bmatrix}$$ 5. **Find the adjugate matrix:** The adjugate is the transpose of the cofactor matrix: $$\mathrm{adj}(A) = C^T = \begin{bmatrix} -3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4 \end{bmatrix}$$ 6. **Calculate the inverse:** $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A) = \frac{1}{-15} \begin{bmatrix} -3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4 \end{bmatrix} = \begin{bmatrix} \frac{-3}{-15} & \frac{5}{-15} & \frac{-2}{-15} \\ \frac{30}{-15} & \frac{-10}{-15} & \frac{-5}{-15} \\ \frac{-24}{-15} & \frac{5}{-15} & \frac{4}{-15} \end{bmatrix}$$ Simplify fractions: $$A^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15} \end{bmatrix}$$ **Final answer:** $$\boxed{A^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15} \end{bmatrix}}$$