1. **Problem Statement:** Find the inverse of the matrices $$A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & 6 & 4 \\ 2 & 4 & -1 \\ -1 & 2 & 5\end{bmatrix}$$ 2. **Step 1: Calculate the determinant of matrix A** $$\det(A) = 1 \times (5 \times 8 - 3 \times 0) - 2 \times (2 \times 8 - 3 \times 1) + 3 \times (2 \times 0 - 5 \times 1)$$ $$= 1 \times (40 - 0) - 2 \times (16 - 3) + 3 \times (0 - 5)$$ $$= 40 - 2 \times 13 - 15 = 40 - 26 - 15 = -1$$ Since $\det(A) \neq 0$, matrix A is invertible. 3. **Step 2: Calculate the inverse of matrix A using the formula** $$A^{-1} = \frac{1}{\det(A)} \times \text{adj}(A)$$ Calculate the matrix of cofactors and then transpose it to get the adjugate matrix: Cofactors of A: $$C = \begin{bmatrix} (5 \times 8 - 3 \times 0) & -(2 \times 8 - 3 \times 1) & (2 \times 0 - 5 \times 1) \\ -(2 \times 8 - 3 \times 1) & (1 \times 8 - 3 \times 1) & -(1 \times 0 - 2 \times 1) \\ (2 \times 0 - 5 \times 1) & -(1 \times 0 - 2 \times 1) & (1 \times 5 - 2 \times 2) \end{bmatrix}$$ $$= \begin{bmatrix} 40 & -13 & -5 \\ -13 & 5 & 2 \\ -5 & 2 & 1 \end{bmatrix}$$ Adjugate matrix $\text{adj}(A)$ is the transpose of $C$: $$\text{adj}(A) = \begin{bmatrix} 40 & -13 & -5 \\ -13 & 5 & 2 \\ -5 & 2 & 1 \end{bmatrix}^T = \begin{bmatrix} 40 & -13 & -5 \\ -13 & 5 & 2 \\ -5 & 2 & 1 \end{bmatrix}$$ 4. **Step 3: Compute $A^{-1}$** $$A^{-1} = \frac{1}{-1} \times \begin{bmatrix} 40 & -13 & -5 \\ -13 & 5 & 2 \\ -5 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -40 & 13 & 5 \\ 13 & -5 & -2 \\ 5 & -2 & -1 \end{bmatrix}$$ 5. **Step 4: Calculate the determinant of matrix B** $$\det(B) = 1 \times (4 \times 5 - (-1) \times 2) - 6 \times (2 \times 5 - (-1) \times (-1)) + 4 \times (2 \times 2 - 4 \times (-1))$$ $$= 1 \times (20 + 2) - 6 \times (10 - 1) + 4 \times (4 + 4)$$ $$= 22 - 6 \times 9 + 4 \times 8 = 22 - 54 + 32 = 0$$ Since $\det(B) = 0$, matrix B is not invertible and does not have an inverse. **Final answers:** $$A^{-1} = \begin{bmatrix} -40 & 13 & 5 \\ 13 & -5 & -2 \\ 5 & -2 & -1 \end{bmatrix}$$ Matrix B has no inverse because its determinant is zero.