Question:

Step 1: Prove that $A$ and $B$ are idempotent matrices. From the given condition $AB = A$, we can write $A = AB$. Multiplying both sides on the right by $A$: $$A^2 = AB \cdot A = A(BA)$$ Since $BA = B$, we substitute: $$A^2 = AB = A$$ Similarly, from $BA = B$, we can write $B = BA$. Multiplying both sides on the right by $B$: $$B^2 = BA \cdot B = B(AB)$$ Since $AB = A$, we substitute: $$B^2 = BA = B$$ Therefore, both $A$ and $B$ are idempotent matrices: $A^2 = A$ and $B^2 = B$. Step 2: Calculate $AB + BA$. From the given conditions: $$AB + BA = A + B$$ Step 3: Expand $(A+B)^2$ using the idempotent property. $$(A+B)^2 = A^2 + AB + BA + B^2$$ Substituting $A^2 = A$, $AB = A$, $BA = B$, and $B^2 = B$: $$(A+B)^2 = A + A + B + B = 2(A+B)$$ Step 4: Establish the general pattern for $(A+B)^n$ using mathematical induction. From Step 3, we have $(A+B)^2 = 2(A+B)$. For $(A+B)^3$: $$(A+B)^3 = (A+B) \cdot (A+B)^2 = (A+B) \cdot 2(A+B) = 2(A+B)^2 = 2 \cdot 2(A+B) = 2^2(A+B)$$ By induction, we can show: $$(A+B)^n = 2^{n-1}(A+B)$$ Step 5: Simplify $A^{2019} + B^{2019}$ using the idempotent property. Since $A$ is idempotent with $A^2 = A$, we have: $$A^{2019} = A$$ Similarly, since $B$ is idempotent with $B^2 = B$, we have: $$B^{2019} = B$$ Therefore: $$A^{2019} + B^{2019} = A + B$$ Step 6: Calculate $(A^{2019} + B^{2019})^{2020}$ using the formula from Step 4. $$(A^{2019} + B^{2019})^{2020} = (A+B)^{2020}$$ Using the formula $(A+B)^n = 2^{n-1}(A+B)$ with $n = 2020$: $$(A+B)^{2020} = 2^{2020-1}(A+B) = 2^{2019}(A+B)$$ **Final Answer:** $(A^{2019} + B^{2019})^{2020} = 2^{2019}(A+B)$ The answer is **Option 3: $2^{2019}(A + B)$**