Question:

<p><strong>Step 1:</strong> Find A² from AB = A.</p><p>Multiply AB = A on the right by B: (AB)B = AB → A(BB) = AB</p><p>Since BA = B, we have: A·B = AB = A (given)</p><p>Also, multiply AB = A on the left by A: A(AB) = AA → A·AB = A²</p><p>Substituting AB = A: A·A = A² → <strong>A² = A</strong></p><p><strong>Step 2:</strong> Similarly prove B² = B.</p><p>From BA = B, multiply by A on right: (BA)A = BA → B(AA) = BA</p><p>Using BA = B and following similar logic: <strong>B² = B</strong></p><p><strong>Step 3:</strong> Determine A^n and B^n.</p><p>Since A² = A, we have A³ = A·A² = A·A = A², and by induction: <strong>A^n = A</strong> for all n ≥ 1</p><p>Similarly: <strong>B^n = B</strong> for all n ≥ 1</p><p><strong>Step 4:</strong> Calculate (A^2019 + B^2019)^2020.</p><p>A^2019 = A and B^2019 = B</p><p>Therefore: (A^2019 + B^2019)^2020 = (A + B)^2020</p><p><strong>Step 5:</strong> Find the value of (A + B)^2020.</p><p>From AB = A and BA = B: AB + BA = A + B</p><p>Note that (A + B)² = A² + AB + BA + B² = A + A + B + B = 2(A + B)</p><p>This means (A + B)² = 2(A + B), so (A + B) is idempotent with factor 2.</p><p>For (A + B)^n: (A + B)^2 = 2(A + B) → (A + B)^2020 = 2^2019(A + B)</p><p>∴ Answer: <strong>D (which represents 2^2019(A + B))</strong></p>