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 AB = A and BA = B, we can derive that A and B are idempotent matrices (A² = A and B² = B) and that AB + BA = A + B. These properties allow us to compute (A + B)ⁿ using the binomial theorem with simplified terms.
<p><strong>Step 1: Find properties of A and B from given conditions.</strong></p><p>Given: AB = A and BA = B</p><p>From AB = A, multiply both sides by B on the right: ABB = AB ⟹ AB² = A</p><p>But AB = A, so AB² = AB ⟹ A(B² - B) = 0</p><p>From BA = B, multiply both sides by A on the right: BAA = BA ⟹ BA² = B</p><p>But BA = B, so BA² = BA ⟹ B(A² - A) = 0</p><p><strong>Step 2: Prove A² = A and B² = B.</strong></p><p>From AB = A: A = AB ⟹ A² = A·AB = (AA)B. Since AB = A, we get A² = AB = A</p><p>From BA = B: B = BA ⟹ B² = B·BA = (BB)A. Since BA = B, we get B² = BA = B</p><p>Thus A and B are idempotent matrices.</p><p><strong>Step 3: Show that A and B commute and find AB + BA.</strong></p><p>Given AB = A and BA = B</p><p>Adding: AB + BA = A + B</p><p><strong>Step 4: Compute (A + B)².</strong></p><p>(A + B)² = A² + AB + BA + B² = A + A + B + B = 2A + 2B = 2(A + B)</p><p><strong>Step 5: Find pattern for (A + B)ⁿ.</strong></p><p>(A + B)³ = (A + B)²(A + B) = 2(A + B)(A + B) = 2(A + B)² = 2·2(A + B) = 4(A + B)</p><p>(A + B)⁴ = (A + B)·(A + B)³ = (A + B)·4(A + B) = 4(A + B)² = 8(A + B)</p><p>By induction: <strong>(A + B)ⁿ = 2^(n-1)(A + B)</strong></p><p><strong>Step 6: Calculate (A + B)⁹.</strong></p><p>(A + B)⁹ = 2^(9-1)(A + B) = 2⁸(A + B) = 256(A + B)</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C