If $A$ and $B$ are two square matrices of order $3 \times 3$ which satisfy $AB = A$ and $BA = B$, then $(A + B)^9$ is equal to
Step-by-Step Solution
Key Concept: From the conditions AB = A and BA = B, we can derive that A and B are idempotent matrices (A² = A and B² = B) and that A + B satisfies a special property. By analyzing the eigenvalues and using the constraint equations, we find that (A + B)² = A + B, making A + B idempotent.
<p><strong>Step 1:</strong> From AB = A, multiply both sides on the right by B: ABB = AB, so AB² = AB = A. This means AB² = A.</p><p><strong>Step 2:</strong> From BA = B, multiply both sides on the left by A: ABA = AB, so ABA = AB = A. This confirms consistency.</p><p><strong>Step 3:</strong> From AB = A, we get A = AB, so A(I - B) = 0. From BA = B, we get B = BA, so (I - A)B = 0.</p><p><strong>Step 4:</strong> Compute (A + B)²: (A + B)² = A² + AB + BA + B² = A² + A + B + B² (using AB = A and BA = B).</p><p><strong>Step 5:</strong> From AB = A, multiply on the left by A: A(AB) = A·A, giving A·A = A². So A² = A (A is idempotent).</p><p><strong>Step 6:</strong> Similarly, from BA = B, multiply on the right by A: (BA)A = B·A, giving BA² = BA = B. Since A² = A, we have B·A = B, confirming BA = B. From BA = B, multiply on the left by B: B(BA) = B·B, giving B² = B (B is idempotent).</p><p><strong>Step 7:</strong> Therefore: (A + B)² = A + A + B + B = 2A + 2B = 2(A + B).</p><p><strong>Step 8:</strong> Let M = A + B. We have M² = 2M. Then M³ = M·M² = M·2M = 2M², and M⁴ = M·M³ = M·2M² = 2M³ = 2·2M² = 4M².</p><p><strong>Step 9:</strong> By induction, if M^n = 2^(n-1)M, then M^(n+1) = M·M^n = M·2^(n-1)M = 2^(n-1)M². Since M² = 2M, we get M^(n+1) = 2^(n-1)·2M = 2^n·M.</p><p><strong>Step 10:</strong> Using M² = 2M as the recurrence, we have: M⁹ = 2^(9-1)·M = 2⁸·M = 256(A + B).</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C