<p><strong>94.</strong> Let <em>A</em> and <em>B</em> are square matrices of same order satisfying \(AB = A\) and \(BA = B\), then \((A^{2019} + B^{2019})^{2020}\) is equal to:</p>
Step-by-Step Solution
Key Concept: From AB = A and BA = B, we can derive that A² = A and B² = B (idempotent matrices), which means A^n = A and B^n = B for all positive integers n.
<p><strong>Step 1:</strong> From AB = A, multiply both sides by B on the right: ABB = AB, so AB² = AB = A</p><p><strong>Step 2:</strong> From BA = B, multiply both sides by A on the right: BAA = BA, so BA² = BA = B</p><p><strong>Step 3:</strong> From AB = A and multiplying by A on the left: A(AB) = A², giving A² = A. Similarly, from BA = B, we get B² = B.</p><p><strong>Step 4:</strong> Since A² = A, we have A^n = A for all positive integers n. Similarly, B^n = B for all positive integers n.</p><p><strong>Step 5:</strong> Therefore, A^2019 = A and B^2019 = B</p><p><strong>Step 6:</strong> So (A^2019 + B^2019)^2020 = (A + B)^2020</p><p><strong>Step 7:</strong> From AB = A and BA = B, we get AB + BA = A + B. Also note that (A + B)² = A² + AB + BA + B² = A + A + B + B = 2A + 2B = 2(A + B)</p><p><strong>Step 8:</strong> By induction, (A + B)^n = 2^(n-1)(A + B) for n ≥ 1</p><p><strong>Step 9:</strong> Therefore, (A + B)^2020 = 2^2019(A + B) = A^2019 + B^2019 raised to appropriate power simplifies to <strong>2^2019(A + B)</strong></p><p>∴ Answer: D</p>
Correct Answer: D