1. **Stating the problem:** We are given two 3x3 matrices: $$A = \begin{bmatrix}4 & 6 & 7 \\ 3 & 4 & 5 \\ 2 & -2 & 3\end{bmatrix}, \quad B = \begin{bmatrix}-3 & 5 & 3 \\ 2 & 3 & 2 \\ 9 & 4 & -6\end{bmatrix}$$ We need to find: a. $A \times B$ b. $B \times A$ c. $(AB)^t$ d. $\det A$ e. $\det B$ f. $\det (AB)$ --- 2. **Matrix multiplication formula:** For matrices $X$ and $Y$, the element at row $i$, column $j$ of $XY$ is: $$ (XY)_{ij} = \sum_{k=1}^n X_{ik} Y_{kj} $$ where $n$ is the number of columns in $X$ (and rows in $Y$). 3. **Calculating $A \times B$:** Calculate each element: - $(AB)_{11} = 4(-3) + 6(2) + 7(9) = -12 + 12 + 63 = 63$ - $(AB)_{12} = 4(5) + 6(3) + 7(4) = 20 + 18 + 28 = 66$ - $(AB)_{13} = 4(3) + 6(2) + 7(-6) = 12 + 12 - 42 = -18$ - $(AB)_{21} = 3(-3) + 4(2) + 5(9) = -9 + 8 + 45 = 44$ - $(AB)_{22} = 3(5) + 4(3) + 5(4) = 15 + 12 + 20 = 47$ - $(AB)_{23} = 3(3) + 4(2) + 5(-6) = 9 + 8 - 30 = -13$ - $(AB)_{31} = 2(-3) + (-2)(2) + 3(9) = -6 - 4 + 27 = 17$ - $(AB)_{32} = 2(5) + (-2)(3) + 3(4) = 10 - 6 + 12 = 16$ - $(AB)_{33} = 2(3) + (-2)(2) + 3(-6) = 6 - 4 - 18 = -16$ So, $$AB = \begin{bmatrix}63 & 66 & -18 \\ 44 & 47 & -13 \\ 17 & 16 & -16\end{bmatrix}$$ --- 4. **Calculating $B \times A$:** - $(BA)_{11} = (-3)(4) + 5(3) + 3(2) = -12 + 15 + 6 = 9$ - $(BA)_{12} = (-3)(6) + 5(4) + 3(-2) = -18 + 20 - 6 = -4$ - $(BA)_{13} = (-3)(7) + 5(5) + 3(3) = -21 + 25 + 9 = 13$ - $(BA)_{21} = 2(4) + 3(3) + 2(2) = 8 + 9 + 4 = 21$ - $(BA)_{22} = 2(6) + 3(4) + 2(-2) = 12 + 12 - 4 = 20$ - $(BA)_{23} = 2(7) + 3(5) + 2(3) = 14 + 15 + 6 = 35$ - $(BA)_{31} = 9(4) + 4(3) + (-6)(2) = 36 + 12 - 12 = 36$ - $(BA)_{32} = 9(6) + 4(4) + (-6)(-2) = 54 + 16 + 12 = 82$ - $(BA)_{33} = 9(7) + 4(5) + (-6)(3) = 63 + 20 - 18 = 65$ So, $$BA = \begin{bmatrix}9 & -4 & 13 \\ 21 & 20 & 35 \\ 36 & 82 & 65\end{bmatrix}$$ --- 5. **Calculating $(AB)^t$ (transpose of $AB$):** Transpose means swapping rows and columns: $$ (AB)^t = \begin{bmatrix}63 & 44 & 17 \\ 66 & 47 & 16 \\ -18 & -13 & -16\end{bmatrix} $$ --- 6. **Calculating $\det A$:** Using cofactor expansion along the first row: $$\det A = 4 \times \det \begin{bmatrix}4 & 5 \\ -2 & 3\end{bmatrix} - 6 \times \det \begin{bmatrix}3 & 5 \\ 2 & 3\end{bmatrix} + 7 \times \det \begin{bmatrix}3 & 4 \\ 2 & -2\end{bmatrix}$$ Calculate minors: - $\det \begin{bmatrix}4 & 5 \\ -2 & 3\end{bmatrix} = 4 \times 3 - 5 \times (-2) = 12 + 10 = 22$ - $\det \begin{bmatrix}3 & 5 \\ 2 & 3\end{bmatrix} = 3 \times 3 - 5 \times 2 = 9 - 10 = -1$ - $\det \begin{bmatrix}3 & 4 \\ 2 & -2\end{bmatrix} = 3 \times (-2) - 4 \times 2 = -6 - 8 = -14$ So, $$\det A = 4(22) - 6(-1) + 7(-14) = 88 + 6 - 98 = -4$$ --- 7. **Calculating $\det B$:** Similarly, expand along the first row: $$\det B = -3 \times \det \begin{bmatrix}3 & 2 \\ 4 & -6\end{bmatrix} - 5 \times \det \begin{bmatrix}2 & 2 \\ 9 & -6\end{bmatrix} + 3 \times \det \begin{bmatrix}2 & 3 \\ 9 & 4\end{bmatrix}$$ Calculate minors: - $\det \begin{bmatrix}3 & 2 \\ 4 & -6\end{bmatrix} = 3 \times (-6) - 2 \times 4 = -18 - 8 = -26$ - $\det \begin{bmatrix}2 & 2 \\ 9 & -6\end{bmatrix} = 2 \times (-6) - 2 \times 9 = -12 - 18 = -30$ - $\det \begin{bmatrix}2 & 3 \\ 9 & 4\end{bmatrix} = 2 \times 4 - 3 \times 9 = 8 - 27 = -19$ So, $$\det B = -3(-26) - 5(-30) + 3(-19) = 78 + 150 - 57 = 171$$ --- 8. **Calculating $\det (AB)$:** Property of determinants: $$\det (AB) = \det A \times \det B$$ So, $$\det (AB) = (-4) \times 171 = -684$$ --- **Final answers:** - a. $A \times B = \begin{bmatrix}63 & 66 & -18 \\ 44 & 47 & -13 \\ 17 & 16 & -16\end{bmatrix}$ - b. $B \times A = \begin{bmatrix}9 & -4 & 13 \\ 21 & 20 & 35 \\ 36 & 82 & 65\end{bmatrix}$ - c. $(AB)^t = \begin{bmatrix}63 & 44 & 17 \\ 66 & 47 & 16 \\ -18 & -13 & -16\end{bmatrix}$ - d. $\det A = -4$ - e. $\det B = 171$ - f. $\det (AB) = -684$