Matrices & Determinants
Matrix equations
Grade 12
Question:
<p>For two \( 3 \times 3 \) matrices \( A \) and \( B \), let \( A + B = 2B' \) and \( 3A + 2B = I_3 \), where \( B' \) is the transpose of \( B \) and \( I_3 \) is \( 3 \times 3 \) identity matrix. Then</p>
<p>\( 10A + 5B = 3I_3 \)</p>
<p>\( 5A + 10B = 2I_3 \)</p>
<p>\( 3A + 6B = 2I_3 \)</p>
<p>\( B + 2A = I_3 \)</p>
Step-by-Step Solution
Key Concept: Set up a system of two matrix equations and solve simultaneously by eliminating one variable. The key is recognizing that B' (transpose of B) can be treated as an unknown and solving the linear system to find explicit forms of A and B.
<p><strong>Step 1:</strong> Write the given equations clearly:</p><p>Equation (1): A + B = 2B'</p><p>Equation (2): 3A + 2B = I₃</p><p><strong>Step 2:</strong> From equation (1), express A in terms of B:</p><p>A = 2B' - B</p><p><strong>Step 3:</strong> Substitute this expression into equation (2):</p><p>3(2B' - B) + 2B = I₃</p><p>6B' - 3B + 2B = I₃</p><p>6B' - B = I₃</p><p><strong>Step 4:</strong> Take transpose of both sides (using (B')' = B and I₃' = I₃):</p><p>6B - B' = I₃</p><p><strong>Step 5:</strong> Now solve the system:</p><p>6B' - B = I₃ ... (i)</p><p>6B - B' = I₃ ... (ii)</p><p>Multiply (i) by 6: 36B' - 6B = 6I₃</p><p>Add to (ii): 36B' - 6B + 6B - B' = 6I₃ + I₃</p><p>35B' = 7I₃</p><p>B' = (1/5)I₃</p><p><strong>Step 6:</strong> Therefore B = (1/5)I₃ (since B = (B')' and if B' is symmetric)</p><p>From 6B' - B = I₃: 6(1/5)I₃ - B = I₃</p><p>B = (6/5)I₃ - I₃ = (1/5)I₃</p><p>From A = 2B' - B: A = 2(1/5)I₃ - (1/5)I₃ = (1/5)I₃</p><p>∴ Answer: D</p>
Correct Answer: D