1. **Problem Statement:** Find the following for matrices $$A=\begin{bmatrix}6 & 2 \\ -4 & 7\end{bmatrix}, B=\begin{bmatrix}-3 & 3 \\ 5 & -4\end{bmatrix}$$ a) The product $AB^T$ b) The adjoint of $A$ c) The inverse of $B$ 2. **Formulae and Rules:** - Transpose of $B$, denoted $B^T$, is obtained by swapping rows and columns. - Matrix multiplication: $(AB)_{ij} = \sum_k A_{ik} B_{kj}$ - Adjoint of $A$ is the transpose of the cofactor matrix of $A$. - Inverse of $B$ is $B^{-1} = \frac{1}{\det(B)} \operatorname{adj}(B)$ if $\det(B) \neq 0$. --- ### a) Calculate $AB^T$ 3. Find $B^T$: $$B^T = \begin{bmatrix}-3 & 5 \\ 3 & -4\end{bmatrix}$$ 4. Multiply $A$ and $B^T$: $$AB^T = \begin{bmatrix}6 & 2 \\ -4 & 7\end{bmatrix} \begin{bmatrix}-3 & 5 \\ 3 & -4\end{bmatrix}$$ Calculate each element: - First row, first column: $$6 \times (-3) + 2 \times 3 = -18 + 6 = -12$$ - First row, second column: $$6 \times 5 + 2 \times (-4) = 30 - 8 = 22$$ - Second row, first column: $$-4 \times (-3) + 7 \times 3 = 12 + 21 = 33$$ - Second row, second column: $$-4 \times 5 + 7 \times (-4) = -20 - 28 = -48$$ So, $$AB^T = \begin{bmatrix}-12 & 22 \\ 33 & -48\end{bmatrix}$$ --- ### b) Find the adjoint of $A$ 5. Calculate the determinant of $A$: $$\det(A) = 6 \times 7 - 2 \times (-4) = 42 + 8 = 50$$ 6. Find the cofactor matrix of $A$: - $C_{11} = + (7) = 7$ - $C_{12} = - (-4) = 4$ - $C_{21} = - (2) = -2$ - $C_{22} = + (6) = 6$ Cofactor matrix: $$\begin{bmatrix}7 & 4 \\ -2 & 6\end{bmatrix}$$ 7. Adjoint is transpose of cofactor matrix: $$\operatorname{adj}(A) = \begin{bmatrix}7 & -2 \\ 4 & 6\end{bmatrix}$$ --- ### c) Find the inverse of $B$ 8. Calculate determinant of $B$: $$\det(B) = (-3)(-4) - 3 \times 5 = 12 - 15 = -3$$ 9. Find cofactor matrix of $B$: - $C_{11} = + (-4) = -4$ - $C_{12} = - (5) = -5$ - $C_{21} = - (3) = -3$ - $C_{22} = + (-3) = -3$ Cofactor matrix: $$\begin{bmatrix}-4 & -5 \\ -3 & -3\end{bmatrix}$$ 10. Adjoint of $B$ is transpose of cofactor matrix: $$\operatorname{adj}(B) = \begin{bmatrix}-4 & -3 \\ -5 & -3\end{bmatrix}$$ 11. Inverse of $B$: $$B^{-1} = \frac{1}{\det(B)} \operatorname{adj}(B) = \frac{1}{-3} \begin{bmatrix}-4 & -3 \\ -5 & -3\end{bmatrix}$$ 12. Simplify: $$B^{-1} = \begin{bmatrix}\frac{-4}{-3} & \frac{-3}{-3} \\ \frac{-5}{-3} & \frac{-3}{-3}\end{bmatrix} = \begin{bmatrix}\frac{4}{3} & 1 \\ \frac{5}{3} & 1\end{bmatrix}$$