1. **Problem Statement:** Given matrices $$A = \begin{bmatrix} 2 & 7 \\ 4 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 9 \\ 1 & 8 \end{bmatrix}$$ Find: (i) $A \cdot B$ (ii) $B \cdot A$ (iii) $A + B$ (iv) $2A - B$ (v) $A - B$ 2. **Matrix Multiplication Formula:** For two matrices $X = [x_{ij}]$ and $Y = [y_{ij}]$, the product $XY$ is defined if the number of columns of $X$ equals the number of rows of $Y$. The element at position $(i,j)$ in $XY$ is: $$ (XY)_{ij} = \sum_k x_{ik} y_{kj} $$ 3. **Matrix Addition and Scalar Multiplication:** - Addition: Add corresponding elements. - Scalar multiplication: Multiply each element by the scalar. 4. **Calculations:** (i) $A \cdot B$: $$A \cdot B = \begin{bmatrix} 2 & 7 \\ 4 & 4 \end{bmatrix} \begin{bmatrix} 4 & 9 \\ 1 & 8 \end{bmatrix} = \begin{bmatrix} (2)(4)+(7)(1) & (2)(9)+(7)(8) \\ (4)(4)+(4)(1) & (4)(9)+(4)(8) \end{bmatrix} = \begin{bmatrix} 8+7 & 18+56 \\ 16+4 & 36+32 \end{bmatrix} = \begin{bmatrix} 15 & 74 \\ 20 & 68 \end{bmatrix}$$ (ii) $B \cdot A$: $$B \cdot A = \begin{bmatrix} 4 & 9 \\ 1 & 8 \end{bmatrix} \begin{bmatrix} 2 & 7 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} (4)(2)+(9)(4) & (4)(7)+(9)(4) \\ (1)(2)+(8)(4) & (1)(7)+(8)(4) \end{bmatrix} = \begin{bmatrix} 8+36 & 28+36 \\ 2+32 & 7+32 \end{bmatrix} = \begin{bmatrix} 44 & 64 \\ 34 & 39 \end{bmatrix}$$ (iii) $A + B$: $$A + B = \begin{bmatrix} 2+4 & 7+9 \\ 4+1 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 16 \\ 5 & 12 \end{bmatrix}$$ (iv) $2A - B$: First calculate $2A$: $$2A = 2 \times \begin{bmatrix} 2 & 7 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 14 \\ 8 & 8 \end{bmatrix}$$ Then subtract $B$: $$2A - B = \begin{bmatrix} 4-4 & 14-9 \\ 8-1 & 8-8 \end{bmatrix} = \begin{bmatrix} 0 & 5 \\ 7 & 0 \end{bmatrix}$$ (v) $A - B$: $$A - B = \begin{bmatrix} 2-4 & 7-9 \\ 4-1 & 4-8 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 3 & -4 \end{bmatrix}$$ **Final answers:** (i) $A \cdot B = \begin{bmatrix} 15 & 74 \\ 20 & 68 \end{bmatrix}$ (ii) $B \cdot A = \begin{bmatrix} 44 & 64 \\ 34 & 39 \end{bmatrix}$ (iii) $A + B = \begin{bmatrix} 6 & 16 \\ 5 & 12 \end{bmatrix}$ (iv) $2A - B = \begin{bmatrix} 0 & 5 \\ 7 & 0 \end{bmatrix}$ (v) $A - B = \begin{bmatrix} -2 & -2 \\ 3 & -4 \end{bmatrix}$