<p>Let <i>A</i> be a square matrix of order 3 such that \(\det(A) = \dfrac{1}{3}\), then the value of \(\det(\text{adj}\, A^{-1})\) is</p>
Step-by-Step Solution
Key Concept: Use the property that det(adj B) = (det B)^(n-1) for an n×n matrix, combined with det(A^(-1)) = 1/det(A) for a 3×3 matrix.
<p><strong>Step 1:</strong> Recall that for an n×n matrix B: det(adj B) = (det B)^(n-1)</p><p><strong>Step 2:</strong> For our 3×3 matrix, det(adj A^(-1)) = (det A^(-1))^(3-1) = (det A^(-1))^2</p><p><strong>Step 3:</strong> Since det(A) = 1/3, we have det(A^(-1)) = 1/det(A) = 1/(1/3) = 3</p><p><strong>Step 4:</strong> Therefore, det(adj A^(-1)) = (3)^2 = 9</p><p>∴ Answer: D (assuming D = 9)</p>
Correct Answer: D