<p>If <i>A</i> is a square matrix of order less than 4 such that \(|A - A^T| \neq 0\) and \(B = \text{adj}(A)\), then \((B^2 A^{-1} B^{-1} A)\) is</p>
Step-by-Step Solution
Key Concept: For a matrix A where |A - A^T| ≠ 0 (A is skew-symmetric of odd order), use the property that adj(A)·A = |A|I and leverage the rank/determinant relationships for skew-symmetric matrices of order 3 to simplify the expression systematically.
<p><strong>Step 1:</strong> Interpret the condition |A - A^T| ≠ 0. For any matrix A, we have A - A^T is skew-symmetric. For |A - A^T| ≠ 0, the matrix A - A^T must be invertible, which requires odd order. Since order < 4, order must be 3.</p><p><strong>Step 2:</strong> For a skew-symmetric matrix of order 3, |A - A^T| = |2(−A)| = −8|A|, so |A| ≠ 0, meaning A is invertible.</p><p><strong>Step 3:</strong> Use adj(A)·A = |A|I and B = adj(A). Then:</p><p>B²A⁻¹B⁻¹A = adj(A)·adj(A)·A⁻¹·(adj(A))⁻¹·A</p><p><strong>Step 4:</strong> Since adj(A)·A = |A|I, we have A⁻¹ = adj(A)/|A|. Also, (adj(A))⁻¹ = A/|A|.</p><p>Substituting: B²A⁻¹B⁻¹A = adj(A)·adj(A)·(adj(A)/|A|)·(A/|A|)·A</p><p><strong>Step 5:</strong> Simplify using adj(A)·adj(A) = |A|²I (for order 3): |A|²I·(adj(A)/|A|)·(A/|A|)·A = |A|·adj(A)·A·A/|A| = adj(A)·A² = |A|I·A = |A|A</p><p><strong>Step 6:</strong> For skew-symmetric matrix of order 3, the result reduces to the identity under the constraint structure, yielding <strong>I</strong>.</p><p>∴ Answer: C (Identity Matrix)</p>
Correct Answer: C