1. We are given the matrix $$A = \begin{bmatrix} 1 & 0 & -2 & -1 \\ 1 & 1 & 1 & 4 \\ 0 & 1 & 2 & 1 \\ -1 & 0 & 3 & 8 \end{bmatrix}$$ and asked to find its determinant $\det A$. 2. The determinant of a $4 \times 4$ matrix can be calculated using cofactor expansion or row operations. 3. Let's use row operations to simplify $A$ to an upper triangular matrix, since the determinant of a triangular matrix is the product of its diagonal entries. 4. Start with $A$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 1 & 1 & 1 & 4 \\ 0 & 1 & 2 & 1 \\ -1 & 0 & 3 & 8 \end{bmatrix}$$ 5. Perform $R_2 \to R_2 - R_1$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 0 & 1 & 3 & 5 \\ 0 & 1 & 2 & 1 \\ -1 & 0 & 3 & 8 \end{bmatrix}$$ 6. Perform $R_4 \to R_4 + R_1$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 0 & 1 & 3 & 5 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 7 \end{bmatrix}$$ 7. Perform $R_3 \to R_3 - R_2$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & -1 & -4 \\ 0 & 0 & 1 & 7 \end{bmatrix}$$ 8. Perform $R_4 \to R_4 + R_3$: $$\begin{bmatrix} 1 & 0 & -2 & -1 \\ 0 & 1 & 3 & 5 \\ 0 & 0 & -1 & -4 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ 9. Now the matrix is upper triangular. The determinant is the product of the diagonal entries: $$\det A = 1 \times 1 \times (-1) \times 3 = -3$$ 10. Therefore, the determinant of $A$ is $-3$. **Final answer:** $\boxed{-3}$