Matrices & Determinants
Properties of Matrices
Grade 12
Question:
<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>Which of the following is true?</p>
<p>If matrix \(A\) is singular, then matrix \(B\) is nonsingular.</p>
<p>If matrix \(A\) is nonsingular, then matrix \(B\) is singular.</p>
<p>If matrix \(A\) is singular, then matrix \(B\) is also singular.</p>
<p>Cannot say anything.</p>
Step-by-Step Solution
Key Concept: From AB = A and BA = B, multiply the first equation by B on the right: ABB = AB, which gives B = A (since AB = A). This reveals that A = B, and both satisfy the idempotent property A² = A.
<p><strong>Step 1:</strong> Given AB = A and BA = B. Multiply the first equation by B on the right side:</p><p>AB · B = A · B</p><p>A(BB) = AB</p><p><strong>Step 2:</strong> From AB = A, substitute on the right:</p><p>AB² = A ... (i)</p><p><strong>Step 3:</strong> Multiply the second equation BA = B by A on the right:</p><p>BA · A = B · A</p><p>B(AA) = BA</p><p>BA² = B ... (ii)</p><p><strong>Step 4:</strong> From BA = B, multiply by A on the left:</p><p>A(BA) = AB</p><p>A · B = A (using BA = B)</p><p>This is already given, so multiply BA = B by B on the right:</p><p>BA · B = B · B</p><p>B² = B² (using AB = A)</p><p><strong>Step 5:</strong> The key insight: From AB = A, we have A(B - I) = 0. From BA = B, we have B(A - I) = 0. These conditions together imply A = B and A² = A.</p><p><strong>Step 6:</strong> Verification: If A = B and A² = A, then AB = A·A = A² = A ✓ and BA = A·A = A² = A = B ✓</p><p>∴ <strong>A and B are equal idempotent matrices satisfying A² = A</strong></p>
Correct Answer: C