The determinant of \( \left[\begin{array}{cccc}-2 & 0 & 0 & 3 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & -3 & -1 \\ -3 & -3 & 0 & 0\end{array}\right] \) is

To find the determinant of the matrix \[ A = \begin{bmatrix} -2 & 0 & 0 & 3 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & -3 & -1 \\ -3 & -3 & 0 & 0 \end{bmatrix}, \] we can use cofactor expansion. It's often easier to expand along a row or column with zeros, and in this case, the first row has two zeros. Let's expand along the first row: \[ \text{det}(A) = -2 \cdot \text{det} \begin{bmatrix} -1 & 2 & 0 \\ 0 & -3 & -1 \\ -3 & 0 & 0 \end{bmatrix} + 3 \cdot \text{det} \begin{bmatrix} 0 & -1 & 2 \\ 0 & 0 & -3 \\ -3 & -3 & 0 \end{bmatrix}. \] Calculating the first determinant: \[ B = \begin{bmatrix} -1 & 2 & 0 \\ 0 & -3 & -1 \\ -3 & 0 & 0 \end{bmatrix}. \] Using the cofactor expansion along the first row: \[ \text{det}(B) = -1 \cdot \text{det} \begin{bmatrix} -3 & -1 \\ 0 & 0 \end{bmatrix} + 2 \cdot \text{det} \begin{bmatrix} 0 & -1 \\ -3 & 0 \end{bmatrix}. \] Calculating these two smaller determinants: 1. The first determinant is \(0\) since its second row is all zeros. 2. The second determinant is \[ \text{det}\begin{bmatrix} 0 & -1 \\ -3 & 0 \end{bmatrix} = (0)(0) - (-1)(-3) = -3. \] Thus, \[ \text{det}(B) = -1(0) + 2(-3) = -6. \] Now returning to the determinant of \(A\): We have \[ \text{det}(A) = -2(-6) + 3 \cdot C, \] where \(C\) is the determinant of \[ \begin{bmatrix} 0 & -1 & 2 \\ 0 & 0 & -3 \\ -3 & -3 & 0 \end{bmatrix}. \] The determinant \(C\) is calculated similarly: \[ C = -1 \cdot \text{det} \begin{bmatrix} 0 & -3 \\ -3 & 0 \end{bmatrix} + 2 \cdot 0 = -1(0 - 9) = 9. \] Finally, plugging \(C\) back into our expression for \(\text{det}(A)\): \[ \text{det}(A) = -2(-6) + 3(9) = 12 + 27 = 39. \] Therefore, the determinant of the matrix is \[ \boxed{39}. \]