1. **State the problem:** We are given matrices $$A = \begin{bmatrix}6 & 0 & 4 \\ 1 & 5 & -3\end{bmatrix}$$ and $$B = \begin{bmatrix}2 & 9 \\ 8 & 0 \\ -4 & 7\end{bmatrix}$$ We need to find: (i) $3A$ (ii) $A \cdot B$ (iii) $B \cdot A$ 2. **Formula and rules:** - Scalar multiplication: multiply each element of the matrix by the scalar. - Matrix multiplication: if $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$ with entries $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$. - Matrix multiplication is only defined if the number of columns of the first matrix equals the number of rows of the second. 3. **Calculate (i) $3A$:** Multiply each element of $A$ by 3: $$3A = \begin{bmatrix}3 \times 6 & 3 \times 0 & 3 \times 4 \\ 3 \times 1 & 3 \times 5 & 3 \times (-3)\end{bmatrix} = \begin{bmatrix}18 & 0 & 12 \\ 3 & 15 & -9\end{bmatrix}$$ 4. **Calculate (ii) $A \cdot B$:** $A$ is $2 \times 3$, $B$ is $3 \times 2$, so product is defined and result is $2 \times 2$. Calculate each element: - Entry (1,1): $6 \times 2 + 0 \times 8 + 4 \times (-4) = 12 + 0 -16 = -4$ - Entry (1,2): $6 \times 9 + 0 \times 0 + 4 \times 7 = 54 + 0 + 28 = 82$ - Entry (2,1): $1 \times 2 + 5 \times 8 + (-3) \times (-4) = 2 + 40 + 12 = 54$ - Entry (2,2): $1 \times 9 + 5 \times 0 + (-3) \times 7 = 9 + 0 - 21 = -12$ So, $$A \cdot B = \begin{bmatrix}-4 & 82 \\ 54 & -12\end{bmatrix}$$ 5. **Calculate (iii) $B \cdot A$:** $B$ is $3 \times 2$, $A$ is $2 \times 3$, so product is defined and result is $3 \times 3$. Calculate each element: - Row 1: - (1,1): $2 \times 6 + 9 \times 1 = 12 + 9 = 21$ - (1,2): $2 \times 0 + 9 \times 5 = 0 + 45 = 45$ - (1,3): $2 \times 4 + 9 \times (-3) = 8 - 27 = -19$ - Row 2: - (2,1): $8 \times 6 + 0 \times 1 = 48 + 0 = 48$ - (2,2): $8 \times 0 + 0 \times 5 = 0 + 0 = 0$ - (2,3): $8 \times 4 + 0 \times (-3) = 32 + 0 = 32$ - Row 3: - (3,1): $-4 \times 6 + 7 \times 1 = -24 + 7 = -17$ - (3,2): $-4 \times 0 + 7 \times 5 = 0 + 35 = 35$ - (3,3): $-4 \times 4 + 7 \times (-3) = -16 - 21 = -37$ So, $$B \cdot A = \begin{bmatrix}21 & 45 & -19 \\ 48 & 0 & 32 \\ -17 & 35 & -37\end{bmatrix}$$ **Final answers:** (i) $3A = \begin{bmatrix}18 & 0 & 12 \\ 3 & 15 & -9\end{bmatrix}$ (ii) $A \cdot B = \begin{bmatrix}-4 & 82 \\ 54 & -12\end{bmatrix}$ (iii) $B \cdot A = \begin{bmatrix}21 & 45 & -19 \\ 48 & 0 & 32 \\ -17 & 35 & -37\end{bmatrix}$