<p>If \(A\) is a \(3 \times 3\) non-singular matrix such that \(AA' = A'A\) and \(B = A^{-1}A'\), then \(BB'\) equals</p>
Step-by-Step Solution
Key Concept: Since AA' = A'A (A commutes with its transpose), we can use this symmetry property combined with the definition B = A⁻¹A' to find BB' by direct multiplication and simplification.
<p><strong>Step 1:</strong> Express B and find B'.</p><p>Given: B = A⁻¹A'</p><p>Then: B' = (A⁻¹A')' = (A')' (A⁻¹)' = A(A⁻¹)' = A(A')⁻¹</p><p><strong>Step 2:</strong> Calculate BB'.</p><p>BB' = (A⁻¹A')[A(A')⁻¹]</p><p>= A⁻¹A'A(A')⁻¹</p><p><strong>Step 3:</strong> Use the commutation property AA' = A'A.</p><p>Since AA' = A'A, we have A'A = AA'</p><p>Therefore: BB' = A⁻¹(A'A)(A')⁻¹ = A⁻¹(AA')(A')⁻¹</p><p><strong>Step 4:</strong> Simplify using associativity.</p><p>BB' = (A⁻¹A)A'(A')⁻¹ = IA'(A')⁻¹ = A'(A')⁻¹ = I</p><p>∴ Answer: D (BB' = I)</p>
Correct Answer: D