1. **Problem Statement:** Given matrices: $$A=\begin{bmatrix}5 & 3\end{bmatrix},\quad B=\begin{bmatrix}4 & -2\\9 & 8\end{bmatrix},\quad C=\begin{bmatrix}3 & 12\\6 & -9\end{bmatrix}$$ Calculate: a) $5A$ b) $3B$ c) $\frac{2}{3}C$ d) $\frac{1}{2}B + \frac{2}{3}C$ And solve for matrix $A$ in the matrix equation: $$2\begin{bmatrix}3 & -2\\5 & 1\end{bmatrix} + A = \frac{1}{3}\begin{bmatrix}12 & 6\\9 & -3\end{bmatrix} - 2A$$ 2. **Calculations for matrix scalar multiplications and additions:** a) Multiply each element of $A$ by 5: $$5A = 5 \times \begin{bmatrix}5 & 3\end{bmatrix} = \begin{bmatrix}5\times 5 & 5\times 3\end{bmatrix} = \begin{bmatrix}25 & 15\end{bmatrix}$$ b) Multiply each element of $B$ by 3: $$3B = 3 \times \begin{bmatrix}4 & -2\\9 & 8\end{bmatrix} = \begin{bmatrix}12 & -6\\27 & 24\end{bmatrix}$$ c) Multiply each element of $C$ by $\frac{2}{3}$: $$\frac{2}{3}C = \frac{2}{3} \times \begin{bmatrix}3 & 12\\6 & -9\end{bmatrix} = \begin{bmatrix}2 & 8\\4 & -6\end{bmatrix}$$ d) Calculate $\frac{1}{2}B + \frac{2}{3}C$: First, $$\frac{1}{2}B = \frac{1}{2} \times \begin{bmatrix}4 & -2\\9 & 8\end{bmatrix} = \begin{bmatrix}2 & -1\\4.5 & 4\end{bmatrix}$$ Then, $$\frac{1}{2}B + \frac{2}{3}C = \begin{bmatrix}2 & -1\\4.5 & 4\end{bmatrix} + \begin{bmatrix}2 & 8\\4 & -6\end{bmatrix} = \begin{bmatrix}4 & 7\\8.5 & -2\end{bmatrix}$$ 3. **Solving for matrix $A$ in the equation:** Given: $$2\begin{bmatrix}3 & -2\\5 & 1\end{bmatrix} + A = \frac{1}{3}\begin{bmatrix}12 & 6\\9 & -3\end{bmatrix} - 2A$$ Step 1: Calculate the scalar multiplications: $$2 \times \begin{bmatrix}3 & -2\\5 & 1\end{bmatrix} = \begin{bmatrix}6 & -4\\10 & 2\end{bmatrix}$$ $$\frac{1}{3} \times \begin{bmatrix}12 & 6\\9 & -3\end{bmatrix} = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}$$ Step 2: Rewrite the equation: $$\begin{bmatrix}6 & -4\\10 & 2\end{bmatrix} + A = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix} - 2A$$ Step 3: Add $2A$ to both sides: $$\begin{bmatrix}6 & -4\\10 & 2\end{bmatrix} + A + 2A = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}$$ $$\begin{bmatrix}6 & -4\\10 & 2\end{bmatrix} + 3A = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}$$ Step 4: Subtract $\begin{bmatrix}6 & -4\\10 & 2\end{bmatrix}$ from both sides: $$3A = \begin{bmatrix}4 & 2\\3 & -1\end{bmatrix} - \begin{bmatrix}6 & -4\\10 & 2\end{bmatrix} = \begin{bmatrix}4-6 & 2-(-4)\\3-10 & -1-2\end{bmatrix} = \begin{bmatrix}-2 & 6\\-7 & -3\end{bmatrix}$$ Step 5: Divide both sides by 3 to solve for $A$: $$A = \frac{1}{3} \times \begin{bmatrix}-2 & 6\\-7 & -3\end{bmatrix} = \begin{bmatrix}-\frac{2}{3} & 2\\-\frac{7}{3} & -1\end{bmatrix}$$ **Final Answers:** - a) $5A = \begin{bmatrix}25 & 15\end{bmatrix}$ - b) $3B = \begin{bmatrix}12 & -6\\27 & 24\end{bmatrix}$ - c) $\frac{2}{3}C = \begin{bmatrix}2 & 8\\4 & -6\end{bmatrix}$ - d) $\frac{1}{2}B + \frac{2}{3}C = \begin{bmatrix}4 & 7\\8.5 & -2\end{bmatrix}$ - Matrix $A$ solved in equation: $$A = \begin{bmatrix}-\frac{2}{3} & 2\\-\frac{7}{3} & -1\end{bmatrix}$$