1. **Problem Statement:** Given matrices $$A=\begin{pmatrix}-1 & 2 \\ 3 & -4\end{pmatrix}, B=\begin{pmatrix}2 & -3 \\ 0 & 1\end{pmatrix}, C=\begin{pmatrix}2 \\ -3\end{pmatrix}$$ Calculate: a) $2A - 3B$ b) $BA$ c) $B^T$ d) $A^{-1}$ e) $AC$ --- 2. **Formulas and Rules:** - Matrix addition/subtraction: add/subtract corresponding elements. - Scalar multiplication: multiply each element by the scalar. - Matrix multiplication: row by column dot product. - Transpose: swap rows and columns. - Inverse of 2x2 matrix $M=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $M^{-1} = \frac{1}{ad-bc} \begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$ if $ad-bc \neq 0$. - Matrix-vector multiplication: multiply matrix rows by vector elements. --- 3. **Calculations:** **a) Calculate $2A - 3B$:** $$2A = 2 \times \begin{pmatrix}-1 & 2 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}-2 & 4 \\ 6 & -8\end{pmatrix}$$ $$3B = 3 \times \begin{pmatrix}2 & -3 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}6 & -9 \\ 0 & 3\end{pmatrix}$$ Now subtract: $$2A - 3B = \begin{pmatrix}-2 & 4 \\ 6 & -8\end{pmatrix} - \begin{pmatrix}6 & -9 \\ 0 & 3\end{pmatrix} = \begin{pmatrix}-2-6 & 4-(-9) \\ 6-0 & -8-3\end{pmatrix} = \begin{pmatrix}-8 & 13 \\ 6 & -11\end{pmatrix}$$ **b) Calculate $BA$:** $$B = \begin{pmatrix}2 & -3 \\ 0 & 1\end{pmatrix}, A = \begin{pmatrix}-1 & 2 \\ 3 & -4\end{pmatrix}$$ Multiply: $$BA = \begin{pmatrix}2 & -3 \\ 0 & 1\end{pmatrix} \begin{pmatrix}-1 & 2 \\ 3 & -4\end{pmatrix} = \begin{pmatrix}2 \times (-1) + (-3) \times 3 & 2 \times 2 + (-3) \times (-4) \\ 0 \times (-1) + 1 \times 3 & 0 \times 2 + 1 \times (-4)\end{pmatrix} = \begin{pmatrix}-2 - 9 & 4 + 12 \\ 0 + 3 & 0 - 4\end{pmatrix} = \begin{pmatrix}-11 & 16 \\ 3 & -4\end{pmatrix}$$ **c) Calculate $B^T$ (transpose of B):** $$B^T = \begin{pmatrix}2 & 0 \\ -3 & 1\end{pmatrix}$$ **d) Calculate $A^{-1}$:** First find determinant: $$\det(A) = (-1)(-4) - (2)(3) = 4 - 6 = -2$$ Since determinant $\neq 0$, inverse exists. $$A^{-1} = \frac{1}{-2} \begin{pmatrix}-4 & -2 \\ -3 & -1\end{pmatrix} = \begin{pmatrix}2 & 1 \\ \frac{3}{2} & \frac{1}{2}\end{pmatrix}$$ **e) Calculate $AC$:** $$A = \begin{pmatrix}-1 & 2 \\ 3 & -4\end{pmatrix}, C = \begin{pmatrix}2 \\ -3\end{pmatrix}$$ Multiply: $$AC = \begin{pmatrix}-1 \times 2 + 2 \times (-3) \\ 3 \times 2 + (-4) \times (-3)\end{pmatrix} = \begin{pmatrix}-2 - 6 \\ 6 + 12\end{pmatrix} = \begin{pmatrix}-8 \\ 18\end{pmatrix}$$ --- **Final answers:** a) $\begin{pmatrix}-8 & 13 \\ 6 & -11\end{pmatrix}$ b) $\begin{pmatrix}-11 & 16 \\ 3 & -4\end{pmatrix}$ c) $\begin{pmatrix}2 & 0 \\ -3 & 1\end{pmatrix}$ d) $\begin{pmatrix}2 & 1 \\ \frac{3}{2} & \frac{1}{2}\end{pmatrix}$ e) $\begin{pmatrix}-8 \\ 18\end{pmatrix}$