1. **Problem Statement:** Solve the following vector and scalar operations: i) $(3 + 2) + (9 + 3)$ ii) $(2xy, 5yg) - (xy, -6y^2)$ iii) $(4, -5) \times (5, -4)$ iv) $(3, 7) \div (3, 2)$ Also, find the additive and multiplicative inverses of: 1) $-7$ 2) $\frac{8}{\sqrt{2}}$ 3) $3$ 4) $0$ --- 2. **Step-by-step solutions:** **i) Scalar addition:** Calculate each sum inside parentheses: $$3 + 2 = 5$$ $$9 + 3 = 12$$ Now add the results: $$5 + 12 = 17$$ **ii) Vector subtraction:** Given vectors $(2xy, 5yg)$ and $(xy, -6y^2)$, subtract component-wise: $$ (2xy - xy, 5yg - (-6y^2)) = (xy, 5yg + 6y^2) $$ **iii) Vector multiplication (dot product):** For vectors $(4, -5)$ and $(5, -4)$, the dot product is: $$4 \times 5 + (-5) \times (-4) = 20 + 20 = 40$$ **iv) Vector division:** Vector division is not defined in standard vector algebra. However, if interpreted as component-wise division: $$\left( \frac{3}{3}, \frac{7}{2} \right) = (1, 3.5)$$ --- **Additive and multiplicative inverses:** 1) For $-7$: - Additive inverse is the number which when added gives zero: $$-7 + 7 = 0$$ So additive inverse is $7$. - Multiplicative inverse is the number which when multiplied gives one: $$-7 \times \left(-\frac{1}{7}\right) = 1$$ So multiplicative inverse is $-\frac{1}{7}$. 2) For $\frac{8}{\sqrt{2}}$: - Additive inverse: $$-\frac{8}{\sqrt{2}}$$ - Multiplicative inverse: $$\frac{\sqrt{2}}{8}$$ 3) For $3$: - Additive inverse: $$-3$$ - Multiplicative inverse: $$\frac{1}{3}$$ 4) For $0$: - Additive inverse: $$0$$ - Multiplicative inverse does not exist because division by zero is undefined. --- **Summary:** - Scalar addition result: $17$ - Vector subtraction result: $(xy, 5yg + 6y^2)$ - Dot product: $40$ - Component-wise division: $(1, 3.5)$ - Additive inverses: $7$, $-\frac{8}{\sqrt{2}}$, $-3$, $0$ - Multiplicative inverses: $-\frac{1}{7}$, $\frac{\sqrt{2}}{8}$, $\frac{1}{3}$, none for $0$