<p>If \(A\) and \(B\) are two matrices of order \(3 \times 3\) where \(|A| = -2\), \(|B| = 2\), then \(|(A^{-1}\text{adj}(B^{-1})\text{adj}(2A^{-1})|\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the properties: |adj(M)| = |M|^(n-1) for n×n matrix, |kM| = k^n|M|, and |M^(-1)| = 1/|M| to systematically evaluate each component before multiplying determinants.
<p><strong>Step 1:</strong> Find |A^(-1)|</p><p>Since |A| = -2, we have |A^(-1)| = 1/|A| = 1/(-2) = -1/2</p><p><strong>Step 2:</strong> Find |adj(B^(-1))|</p><p>First, |B^(-1)| = 1/|B| = 1/2</p><p>For a 3×3 matrix M: |adj(M)| = |M|^(3-1) = |M|^2</p><p>Therefore, |adj(B^(-1))| = |B^(-1)|^2 = (1/2)^2 = 1/4</p><p><strong>Step 3:</strong> Find |adj(2A^(-1))|</p><p>First, |2A^(-1)| = 2^3 · |A^(-1)| = 8 · (-1/2) = -4</p><p>Therefore, |adj(2A^(-1))| = |2A^(-1)|^2 = (-4)^2 = 16</p><p><strong>Step 4:</strong> Find the final determinant</p><p>|(A^(-1)·adj(B^(-1))·adj(2A^(-1)))| = |A^(-1)| · |adj(B^(-1))| · |adj(2A^(-1))|</p><p>= (-1/2) · (1/4) · 16 = -1/2 · 4 = <strong>-2</strong></p><p>∴ Answer: <strong>-2</strong></p>
Correct Answer: -2