<p><strong>For Problems 4–6</strong><br>If \(A\) and \(B\) are two square matrices of order \(3 \times 3\) which satisfy \(AB = A\) and \(BA = B\), then<br><br>\((A + B)^7\) 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). This means A and B are projection matrices, so (A + B)⁷ can be simplified using the binomial expansion with careful tracking of which cross terms survive.
<p><strong>Step 1:</strong> From AB = A, multiply both sides on the right by B: ABB = AB → AB² = A. But from BA = B, we have AB = A, so AB² = A.</p><p><strong>Step 2:</strong> Similarly, from BA = B, multiply on the left by A: ABA = AB → A²B = A. From AB = A, we get A²B = A. Also, multiply BA = B on the right by A: BAA = BA → BA² = B, so A² = A (applying BA = B).</p><p><strong>Step 3:</strong> From AB = A and BA = B, we can verify: A² = A(BA) = (AB)A = AA = A². Also B² = B(AB) = (BA)B = BB = B². So both A and B are idempotent.</p><p><strong>Step 4:</strong> Now compute (A + B)²: (A + B)² = A² + AB + BA + B² = A + A + B + B = 2A + 2B = 2(A + B).</p><p><strong>Step 5:</strong> This means (A + B)² = 2(A + B). Let C = A + B. Then C² = 2C, so C³ = C·C² = C·2C = 2C² = 2(2C) = 4C = 2²C.</p><p><strong>Step 6:</strong> By induction: Cⁿ = 2ⁿ⁻¹C for n ≥ 1. Therefore: (A + B)⁷ = C⁷ = 2⁶(A + B) = <strong>64(A + B)</strong></p><p>∴ Answer: D</p>
Correct Answer: D