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 $$A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$$ where $\operatorname{adj}(A)$ 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: $$= 1 \times (4 \times 8 - 5 \times 7) - 2 \times (0 \times 8 - 5 \times 6) + 3 \times (0 \times 7 - 4 \times 6)$$ $$= 1 \times (32 - 35) - 2 \times (0 - 30) + 3 \times (0 - 24)$$ $$= 1 \times (-3) - 2 \times (-30) + 3 \times (-24)$$ $$= -3 + 60 - 72 = -15$$ 4. **Since $\det(A) = -15 \neq 0$, the inverse exists.** 5. **Find the matrix of cofactors:** $$C = \begin{bmatrix} +\begin{vmatrix}4 & 5 \\ 7 & 8\end{vmatrix} & -\begin{vmatrix}0 & 5 \\ 6 & 8\end{vmatrix} & +\begin{vmatrix}0 & 4 \\ 6 & 7\end{vmatrix} \\ -\begin{vmatrix}2 & 3 \\ 7 & 8\end{vmatrix} & +\begin{vmatrix}1 & 3 \\ 6 & 8\end{vmatrix} & -\begin{vmatrix}1 & 2 \\ 6 & 7\end{vmatrix} \\ +\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} & -\begin{vmatrix}1 & 3 \\ 0 & 5\end{vmatrix} & +\begin{vmatrix}1 & 2 \\ 0 & 4\end{vmatrix} \end{bmatrix}$$ Calculate each minor: $$C = \begin{bmatrix} (4 \times 8 - 5 \times 7) & -(0 \times 8 - 5 \times 6) & (0 \times 7 - 4 \times 6) \\ -(2 \times 8 - 3 \times 7) & (1 \times 8 - 3 \times 6) & -(1 \times 7 - 2 \times 6) \\ (2 \times 5 - 3 \times 4) & -(1 \times 5 - 3 \times 0) & (1 \times 4 - 2 \times 0) \end{bmatrix}$$ $$= \begin{bmatrix} (32 - 35) & -(0 - 30) & (0 - 24) \\ -(16 - 21) & (8 - 18) & -(7 - 12) \\ (10 - 12) & -(5 - 0) & (4 - 0) \end{bmatrix}$$ $$= \begin{bmatrix} -3 & 30 & -24 \\ 5 & -10 & -(-5) \\ -2 & -5 & 4 \end{bmatrix} = \begin{bmatrix} -3 & 30 & -24 \\ 5 & -10 & 5 \\ -2 & -5 & 4 \end{bmatrix}$$ 6. **Find the adjugate matrix by transposing the cofactor matrix:** $$\operatorname{adj}(A) = C^T = \begin{bmatrix} -3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4 \end{bmatrix}$$ 7. **Calculate the inverse:** $$A^{-1} = \frac{1}{-15} \begin{bmatrix} -3 & 5 & -2 \\ 30 & -10 & -5 \\ -24 & 5 & 4 \end{bmatrix} = \begin{bmatrix} \frac{-3}{-15} & \frac{5}{-15} & \frac{-2}{-15} \\ \frac{30}{-15} & \frac{-10}{-15} & \frac{-5}{-15} \\ \frac{-24}{-15} & \frac{5}{-15} & \frac{4}{-15} \end{bmatrix}$$ Simplify fractions: $$= \begin{bmatrix} \frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15} \end{bmatrix}$$ **Final answer:** $$A^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{1}{3} & \frac{2}{15} \\ -2 & \frac{2}{3} & \frac{1}{3} \\ \frac{8}{5} & -\frac{1}{3} & -\frac{4}{15} \end{bmatrix}$$