1. **Problem A1:** Given a square matrix $A$ such that $A^3 = I$, determine which statement must be true. 2. **Recall:** The determinant of a product of matrices equals the product of their determinants, so $|A^3| = |A|^3$. 3. Since $A^3 = I$, taking determinants on both sides gives: $$|A^3| = |I|$$ $$|A|^3 = 1$$ 4. This implies: $$|A| = oot 3 elax 1 = 1$$ 5. Check the options: - (a) $|A| = \frac{1}{3}$ is false. - (b) $|A| = 3$ is false. - (c) $A^2 = A^{-1}$ is true because multiplying both sides of $A^3 = I$ by $A^{-1}$ gives $A^2 = A^{-1}$. - (d) $A^2 = I$ is not necessarily true. **Answer for A1:** (c) $A^2 = A^{-1}$ --- 1. **Problem A2:** Given two square matrices $A$ and $B$ where $AB = BA$, determine which statements are always true: - Equation I: $A^{-1} = B$ - Equation II: $(A + B)(A - B) = A^2 - B^2$ 2. **Check Equation I:** $A^{-1} = B$ is not necessarily true just because $A$ and $B$ commute. 3. **Check Equation II:** Since $A$ and $B$ commute, the distributive property applies: $$(A + B)(A - B) = A^2 - AB + BA - B^2$$ Because $AB = BA$, the middle terms cancel: $$-AB + BA = 0$$ So: $$(A + B)(A - B) = A^2 - B^2$$ 4. **Answer for A2:** (b) Equation II only --- **Final answers:** - For A1: (c) $A^2 = A^{-1}$ - For A2: (b) Equation II only