<p>If <em>A</em> is an 3 × 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), the matrix A is normal. Use this property along with B = A⁻¹A' to find BB' by computing B·B' and simplifying using the commutativity condition.
<p><strong>Step 1:</strong> Find B'.</p><p>B = A⁻¹A', so B' = (A⁻¹A')' = (A')ᵀ(A⁻¹)ᵀ = A(A⁻¹)ᵀ</p><p><strong>Step 2:</strong> Compute BB'.</p><p>BB' = (A⁻¹A') · A(A⁻¹)ᵀ = A⁻¹A'A(A⁻¹)ᵀ</p><p><strong>Step 3:</strong> Apply the condition AA' = A'A (given).</p><p>Since AA' = A'A, we have A'A = AA'. Therefore:</p><p>BB' = A⁻¹(AA')(A⁻¹)ᵀ = A⁻¹AA'(A⁻¹)ᵀ = I·A'(A⁻¹)ᵀ</p><p><strong>Step 4:</strong> Simplify using (A⁻¹)ᵀ = (Aᵀ)⁻¹.</p><p>BB' = A'(A')⁻¹ = I</p><p>∴ Answer: <strong>B (I or Identity matrix)</strong></p>
Correct Answer: B