4.) Determine the determinant of the matrix \( \left[\begin{array}{cccc}1 & -1 & 8 & 4 \\ 2 & 6 & 0 & -4 \\ 2 & 0 & 2 & 6 \\ 0 & 2 & 8 & 0\end{array}\right] \).

To find the determinant of the matrix \[ A = \begin{bmatrix} 1 & -1 & 8 & 4 \\ 2 & 6 & 0 & -4 \\ 2 & 0 & 2 & 6 \\ 0 & 2 & 8 & 0 \end{bmatrix}, \] we can use the method of cofactor expansion along the first row or any convenient row or column. Here, I'll expand along the first row for clarity: \[ \text{det}(A) = 1 \cdot M_{11} - (-1) \cdot M_{12} + 8 \cdot M_{13} + 4 \cdot M_{14}, \] where \(M_{ij}\) is the minor of element \(a_{ij}\). Let's compute the minors: 1. **Minor \(M_{11}\)** (remove first row and first column): \[ M_{11} = \begin{vmatrix} 6 & 0 & -4 \\ 0 & 2 & 6 \\ 2 & 8 & 0 \end{vmatrix}. \] Expanding along the first row: \[ = 6 \begin{vmatrix} 2 & 6 \\ 8 & 0 \end{vmatrix} + 0 -4 \begin{vmatrix} 0 & 2 \\ 2 & 8 \end{vmatrix}. \] Finding each 2x2 determinant: \[ \begin{vmatrix} 2 & 6 \\ 8 & 0 \end{vmatrix} = 2 \cdot 0 - 6 \cdot 8 = -48, \] \[ \begin{vmatrix} 0 & 2 \\ 2 & 8 \end{vmatrix} = 0 \cdot 8 - 2 \cdot 2 = -4. \] So: \[ M_{11} = 6(-48) + 0 -4(-4) = -288 + 16 = -272. \] 2. **Minor \(M_{12}\)** (remove first row and second column): \[ M_{12} = \begin{vmatrix} 2 & 0 & -4 \\ 2 & 2 & 6 \\ 0 & 8 & 0 \end{vmatrix}. \] Expanding along the first row: \[ = 2 \begin{vmatrix} 2 & 6 \\ 8 & 0 \end{vmatrix} - 0 + 4 \begin{vmatrix} 2 & 2 \\ 0 & 8 \end{vmatrix}. \] We already calculated \( \begin{vmatrix} 2 & 6 \\ 8 & 0 \end{vmatrix} = -48\), so: \[ = 2(-48) + 4(2 \cdot 8 - 2 \cdot 0) = -96 + 64 = -32. \] 3. **Minor \(M_{13}\)** (remove first row and third column): \[ M_{13} = \begin{vmatrix} 2 & 6 & -4 \\ 2 & 0 & 6 \\ 0 & 2 & 0 \end{vmatrix}. \] Expanding along the first row: \[ = 2 \begin{vmatrix} 0 & 6 \\ 2 & 0 \end{vmatrix} - 6 \begin{vmatrix} 2 & -4 \\ 0 & 0 \end{vmatrix} - 4 \begin{vmatrix} 2 & 6 \\ 0 & 2 \end{vmatrix}. \] Finding the determinant: \[ \begin{vmatrix} 0 & 6 \\ 2 & 0 \end{vmatrix} = 0 \cdot 0 - 6 \cdot 2 = -12, \] \[ \begin{vmatrix} 2 & -4 \\ 0 & 0 \end