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 :
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- \(5A + 10B = 2I_3\)
- \(10A + 5B = 3I_3\)
- \(B + 2A = I_3\)
- \(3A + 6B = 2I_3\)
The Correct Option is B
Solution and Explanation
Transpose (i): \(A' + B' = 2B\). Using \(B' = \frac{A+B}{2}\) from (i): \(2A' + A + B = 4B \Rightarrow 2A' = 3B - A\) ... (iii).
Transpose (ii): \(3A' + 2B' = I\), i.e. \(3A' + A + B = I\) ... (iv).
Put \(A' = \frac{3B - A}{2}\) from (iii) in (iv): \(\frac{9B - 3A}{2} + A + B = I \Rightarrow 11B - A = 2I\), so \(A = 11B - 2I\).
Substitute in (ii): \(33B - 6I + 2B = I \Rightarrow B = \frac{I}{5}\), and \(A = \frac{11}{5}I - 2I = \frac{I}{5}\).
Check: \(A + B = \frac{2}{5}I = 2B'\) holds.
\(10A + 5B = 2I + I = 3I_3\). The other options give \(5A + 10B = 3I\), \(B + 2A = \frac35 I\), \(3A + 6B = \frac95 I\).
Answer: (2) \(10A + 5B = 3I_3\)
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