1. **Problem 1: Find the determinant of matrix A** Matrix A is given by: $$A = \begin{pmatrix} 2 & -1 & 4 \\ -4 & 3 & 0 \\ 5 & -2 & 1 \end{pmatrix}$$ 2. To find $\det(A)$, use the cofactor expansion along the first row: $$\det(A) = 2 \cdot \det\begin{pmatrix} 3 & 0 \\ -2 & 1 \end{pmatrix} - (-1) \cdot \det\begin{pmatrix} -4 & 0 \\ 5 & 1 \end{pmatrix} + 4 \cdot \det\begin{pmatrix} -4 & 3 \\ 5 & -2 \end{pmatrix}$$ 3. Calculate each 2x2 determinant: $$\det\begin{pmatrix} 3 & 0 \\ -2 & 1 \end{pmatrix} = (3)(1) - (0)(-2) = 3$$ $$\det\begin{pmatrix} -4 & 0 \\ 5 & 1 \end{pmatrix} = (-4)(1) - (0)(5) = -4$$ $$\det\begin{pmatrix} -4 & 3 \\ 5 & -2 \end{pmatrix} = (-4)(-2) - (3)(5) = 8 - 15 = -7$$ 4. Substitute back: $$\det(A) = 2 \times 3 + 1 \times (-4) + 4 \times (-7) = 6 - 4 - 28 = -26$$ --- 5. **Problem 2: Find $p$ such that $|B| = 6$** Matrix B is: $$B = \begin{pmatrix} 1 & -4 & -2 \\ 3 & 2 & 2 \\ 0 & p & 1 \end{pmatrix}$$ 6. Calculate $\det(B)$ using cofactor expansion along the first row: $$\det(B) = 1 \cdot \det\begin{pmatrix} 2 & 2 \\ p & 1 \end{pmatrix} - (-4) \cdot \det\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} + (-2) \cdot \det\begin{pmatrix} 3 & 2 \\ 0 & p \end{pmatrix}$$ 7. Calculate each 2x2 determinant: $$\det\begin{pmatrix} 2 & 2 \\ p & 1 \end{pmatrix} = 2 \times 1 - 2 \times p = 2 - 2p$$ $$\det\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} = 3 \times 1 - 2 \times 0 = 3$$ $$\det\begin{pmatrix} 3 & 2 \\ 0 & p \end{pmatrix} = 3 \times p - 2 \times 0 = 3p$$ 8. Substitute back: $$\det(B) = 1 \times (2 - 2p) + 4 \times 3 - 2 \times 3p = (2 - 2p) + 12 - 6p = 14 - 8p$$ 9. Set $\det(B) = 6$ and solve for $p$: $$14 - 8p = 6$$ $$-8p = 6 - 14 = -8$$ $$p = \frac{-8}{-8} = 1$$ **Final answers:** - $\det(A) = -26$ - $p = 1$ so that $|B| = 6$