<p>If <i>A</i> and <i>B</i> are two non-singular matrices of order 3 such that \(AA^T = 2I\) and \(A^{-1} = A^T - A \cdot \text{adj}(2B^{-1})\), then \(\det(B)\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the constraint AA^T = 2I to find det(A) = ±√8, then substitute A^(-1) = A^T into the given equation and compare coefficients to isolate adj(2B^(-1)), finally extract det(B) using properties of adjugate matrices.
<p><strong>Step 1:</strong> From AA^T = 2I, taking determinants: det(A)·det(A^T) = det(2I) = 2³ = 8</p><p>Since det(A^T) = det(A), we get [det(A)]² = 8, so det(A) = ±2√2</p><p><strong>Step 2:</strong> From AA^T = 2I, multiply both sides by A^(-1) on the left: A^T = 2A^(-1)</p><p>Therefore: A^(-1) = (1/2)A^T</p><p><strong>Step 3:</strong> Substitute into the given equation: (1/2)A^T = A^T - A·adj(2B^(-1))</p><p>Simplifying: A·adj(2B^(-1)) = (1/2)A^T</p><p><strong>Step 4:</strong> Multiply both sides by A^T on the right: A·adj(2B^(-1))·A^T = (1/2)A^T·A^T = (1/2)(A^T)²</p><p>Since A^T = 2A^(-1), we get: A·adj(2B^(-1))·A^T = 2I</p><p><strong>Step 5:</strong> Taking determinants: det(A)·det(adj(2B^(-1)))·det(A^T) = det(2I) = 8</p><p>[det(A)]²·det(adj(2B^(-1))) = 8</p><p><strong>Step 6:</strong> Using det(adj(2B^(-1))) = [det(2B^(-1))]² = (2/det(B))²</p><p>We have: 8·(2/det(B))² = 8</p><p>Therefore: (2/det(B))² = 1, so det(B) = ±2</p><p>∴ Answer: D</p>
Correct Answer: D