1. **Problem Statement:** 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 for matrix multiplication:** The element in the $i^{th}$ row and $j^{th}$ column of the product matrix $AB$ is given by $$ (AB)_{ij} = \sum_{k=1}^n A_{ik} B_{kj} $$ where $n$ is the number of columns in $A$ (or rows in $B$). 3. **Calculate each element of $AB$: ** - First row, first column: $$2\times1 + (-3)\times7 + 4\times1 = 2 - 21 + 4 = -15$$ - First row, second column: $$2\times3 + (-3)\times(-3) + 4\times0 = 6 + 9 + 0 = 15$$ - First row, third column: $$2\times(-5) + (-3)\times2 + 4\times(-8) = -10 - 6 - 32 = -48$$ - Second row, first column: $$1\times1 + (-6)\times7 + 3\times1 = 1 - 42 + 3 = -38$$ - Second row, second column: $$1\times3 + (-6)\times(-3) + 3\times0 = 3 + 18 + 0 = 21$$ - Second row, third column: $$1\times(-5) + (-6)\times2 + 3\times(-8) = -5 - 12 - 24 = -41$$ - Third row, first column: $$0\times1 + 5\times7 + (-9)\times1 = 0 + 35 - 9 = 26$$ - Third row, second column: $$0\times3 + 5\times(-3) + (-9)\times0 = 0 - 15 + 0 = -15$$ - Third row, third column: $$0\times(-5) + 5\times2 + (-9)\times(-8) = 0 + 10 + 72 = 82$$ 4. **Result matrix $AB$ is:** $$\begin{pmatrix}-15 & 15 & -48 \\ -38 & 21 & -41 \\ 26 & -15 & 82\end{pmatrix}$$ 5. **Answer:** Option (b) matches the calculated matrix. This completes the solution for the first question.