<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>\((A + I)^5\) is equal to (where \(I\) is identity matrix)</p>
Step-by-Step Solution
Key Concept: From AB = A and BA = B, we can deduce that A and B are idempotent projections satisfying A² = A and B² = B, which means A and B commute and A = B. This allows us to express (A + I)⁵ using the binomial theorem with A² = A.
<p><strong>Step 1:</strong> From AB = A, multiply both sides by B on the right: ABB = AB, so AB² = A. But BA = B, so AB = BA (they commute).</p><p><strong>Step 2:</strong> From AB = A and BA = B, we get A = AB = BAB = B(BA) = BB = B². Also, from BA = B, we have B = BA = ABA = A(AB) = AA = A². Therefore A² = A and B² = B, and A = B.</p><p><strong>Step 3:</strong> Since A² = A, we can compute powers of (A + I):<br/>(A + I)² = A² + 2A + I = A + 2A + I = 3A + I<br/>(A + I)³ = (A + I)(3A + I) = 3A² + A + 3A + I = 3A + 4A + I = 7A + I<br/>(A + I)⁴ = (A + I)(7A + I) = 7A² + A + 7A + I = 7A + 8A + I = 15A + I<br/>(A + I)⁵ = (A + I)(15A + I) = 15A² + A + 15A + I = 15A + 16A + I = 31A + I</p><p><strong>Step 4:</strong> Pattern recognition: (A + I)ⁿ = (2ⁿ - 1)A + I</p><p>∴ Answer: <strong>(A + I)⁵ = 31A + I</strong></p>
Correct Answer: C