1. The problem is to find the inverse of the matrix $$A = \begin{pmatrix} -1 & -3 & -1 \\ 0 & 0 & -2 \\ -1 & 1 & 1 \end{pmatrix}$$. 2. To find the inverse of a 3x3 matrix, we use the formula: $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ where $\det(A)$ is the determinant of $A$ and $\mathrm{adj}(A)$ is the adjugate matrix of $A$. 3. First, calculate the determinant $\det(A)$: $$\det(A) = -1 \times \begin{vmatrix} 0 & -2 \\ 1 & 1 \end{vmatrix} - (-3) \times \begin{vmatrix} 0 & -2 \\ -1 & 1 \end{vmatrix} + (-1) \times \begin{vmatrix} 0 & 0 \\ -1 & 1 \end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix} 0 & -2 \\ 1 & 1 \end{vmatrix} = 0 \times 1 - (-2) \times 1 = 2$$ $$\begin{vmatrix} 0 & -2 \\ -1 & 1 \end{vmatrix} = 0 \times 1 - (-2) \times (-1) = -2$$ $$\begin{vmatrix} 0 & 0 \\ -1 & 1 \end{vmatrix} = 0 \times 1 - 0 \times (-1) = 0$$ So, $$\det(A) = -1 \times 2 - (-3) \times (-2) + (-1) \times 0 = -2 - 6 + 0 = -8$$ 4. Since $\det(A) \neq 0$, the inverse exists. 5. Next, find the matrix of cofactors $C$ by calculating the cofactor for each element: For example, cofactor $C_{11} = (+1) \times \begin{vmatrix} 0 & -2 \\ 1 & 1 \end{vmatrix} = 2$ Calculate all cofactors: $$C = \begin{pmatrix} 2 & 2 & 0 \\ -2 & 0 & -2 \\ 0 & -2 & 0 \end{pmatrix}$$ 6. The adjugate matrix $\mathrm{adj}(A)$ is the transpose of the cofactor matrix: $$\mathrm{adj}(A) = C^T = \begin{pmatrix} 2 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & -2 & 0 \end{pmatrix}$$ 7. Finally, compute the inverse: $$A^{-1} = \frac{1}{-8} \begin{pmatrix} 2 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & -2 & 0 \end{pmatrix} = \begin{pmatrix} -\frac{1}{4} & \frac{1}{4} & 0 \\ -\frac{1}{4} & 0 & \frac{1}{4} \\ 0 & \frac{1}{4} & 0 \end{pmatrix}$$ This is the inverse of matrix $A$.