1. **Problem:** Find the product of matrices $A$ and $B$ where $$A=\begin{pmatrix}2 & -3 & 4 \\ 1 & -6 & 3 \\ 0 & 5 & -9\end{pmatrix},\quad B=\begin{pmatrix}1 & 3 & -5 \\ 7 & -3 & 2 \\ 1 & 0 & -8\end{pmatrix}$$ 2. **Formula:** The product $AB$ is calculated by multiplying rows of $A$ by columns of $B$: $$ (AB)_{ij} = \sum_{k=1}^3 A_{ik} B_{kj} $$ 3. **Calculate each element:** - First row: - $(AB)_{11} = 2\times1 + (-3)\times7 + 4\times1 = 2 - 21 + 4 = -15$ - $(AB)_{12} = 2\times3 + (-3)\times(-3) + 4\times0 = 6 + 9 + 0 = 15$ - $(AB)_{13} = 2\times(-5) + (-3)\times2 + 4\times(-8) = -10 - 6 - 32 = -48$ - Second row: - $(AB)_{21} = 1\times1 + (-6)\times7 + 3\times1 = 1 - 42 + 3 = -38$ - $(AB)_{22} = 1\times3 + (-6)\times(-3) + 3\times0 = 3 + 18 + 0 = 21$ - $(AB)_{23} = 1\times(-5) + (-6)\times2 + 3\times(-8) = -5 - 12 - 24 = -41$ - Third row: - $(AB)_{31} = 0\times1 + 5\times7 + (-9)\times1 = 0 + 35 - 9 = 26$ - $(AB)_{32} = 0\times3 + 5\times(-3) + (-9)\times0 = 0 - 15 + 0 = -15$ - $(AB)_{33} = 0\times(-5) + 5\times2 + (-9)\times(-8) = 0 + 10 + 72 = 82$ 4. **Result:** $$AB = \begin{pmatrix}-15 & 15 & -48 \\ -38 & 21 & -41 \\ 26 & -15 & 82\end{pmatrix}$$ 5. **Answer:** Option (b) matches the computed matrix. --- Since the user asked multiple questions, but per instructions, only the first question is solved here.