1. **Problem Statement:** Find the specified rows and columns of the matrix products $AB$, $BA$, and $AA$ where $$A=\begin{bmatrix}3 & -2 & 7 \\ 6 & 5 & 4 \\ 0 & 4 & 9\end{bmatrix},\quad B=\begin{bmatrix}6 & -2 & 4 \\ 0 & 1 & 3 \\ 7 & 7 & 5\end{bmatrix}$$ 2. **Matrix Multiplication Reminder:** The element in row $i$, column $j$ of the product $XY$ is computed as the dot product of the $i$th row of $X$ and the $j$th column of $Y$: $$ (XY)_{ij} = \sum_{k} X_{ik} Y_{kj} $$ 3. **Calculate $AB$:** - First row of $AB$: $$\begin{aligned} (AB)_{1,1} &= 3\cdot6 + (-2)\cdot0 + 7\cdot7 = 18 + 0 + 49 = 67 \\ (AB)_{1,2} &= 3\cdot(-2) + (-2)\cdot1 + 7\cdot7 = -6 - 2 + 49 = 41 \\ (AB)_{1,3} &= 3\cdot4 + (-2)\cdot3 + 7\cdot5 = 12 - 6 + 35 = 41 \end{aligned}$$ So first row of $AB$ is $[67, 41, 41]$. - Third row of $AB$: $$\begin{aligned} (AB)_{3,1} &= 0\cdot6 + 4\cdot0 + 9\cdot7 = 0 + 0 + 63 = 63 \\ (AB)_{3,2} &= 0\cdot(-2) + 4\cdot1 + 9\cdot7 = 0 + 4 + 63 = 67 \\ (AB)_{3,3} &= 0\cdot4 + 4\cdot3 + 9\cdot5 = 0 + 12 + 45 = 57 \end{aligned}$$ So third row of $AB$ is $[63, 67, 57]$. - Second column of $AB$ (dot product of each row of $A$ with second column of $B$): $$\begin{aligned} (AB)_{1,2} &= 41 \text{ (from above)} \\ (AB)_{2,2} &= 6\cdot(-2) + 5\cdot1 + 4\cdot7 = -12 + 5 + 28 = 21 \\ (AB)_{3,2} &= 67 \text{ (from above)} \end{aligned}$$ So second column of $AB$ is $[41, 21, 67]$. 4. **Calculate $BA$:** - First column of $BA$ (dot product of each row of $B$ with first column of $A$): $$\begin{aligned} (BA)_{1,1} &= 6\cdot3 + (-2)\cdot6 + 4\cdot0 = 18 - 12 + 0 = 6 \\ (BA)_{2,1} &= 0\cdot3 + 1\cdot6 + 3\cdot0 = 0 + 6 + 0 = 6 \\ (BA)_{3,1} &= 7\cdot3 + 7\cdot6 + 5\cdot0 = 21 + 42 + 0 = 63 \end{aligned}$$ So first column of $BA$ is $[6, 6, 63]$. 5. **Calculate $AA$:** - Third row of $AA$: $$\begin{aligned} (AA)_{3,1} &= 0\cdot3 + 4\cdot6 + 9\cdot0 = 0 + 24 + 0 = 24 \\ (AA)_{3,2} &= 0\cdot(-2) + 4\cdot5 + 9\cdot4 = 0 + 20 + 36 = 56 \\ (AA)_{3,3} &= 0\cdot7 + 4\cdot4 + 9\cdot9 = 0 + 16 + 81 = 97 \end{aligned}$$ So third row of $AA$ is $[24, 56, 97]$. - Third column of $AA$ (dot product of each row of $A$ with third column of $A$): $$\begin{aligned} (AA)_{1,3} &= 3\cdot7 + (-2)\cdot4 + 7\cdot9 = 21 - 8 + 63 = 76 \\ (AA)_{2,3} &= 6\cdot7 + 5\cdot4 + 4\cdot9 = 42 + 20 + 36 = 98 \\ (AA)_{3,3} &= 97 \text{ (from above)} \end{aligned}$$ So third column of $AA$ is $[76, 98, 97]$. **Final answers:** - a. First row of $AB$ is $[67, 41, 41]$ - b. Third row of $AB$ is $[63, 67, 57]$ - c. Second column of $AB$ is $[41, 21, 67]$ - d. First column of $BA$ is $[6, 6, 63]$ - e. Third row of $AA$ is $[24, 56, 97]$ - f. Third column of $AA$ is $[76, 98, 97]$