Let A be a square matrix of order 3 such that det (A) > 0 and det ( adj (4 adj (2A))) = 2³² . Then the value of det (4A⁻¹) is equal to _______

Step-by-step solution

Let |A| denote the determinant of matrix A . We know that for a square matrix M of order n , det ( adj (M)) = ( det (M))^ n-1 and det (kM) = k^n det (M) . Here, n=3 . Let X = 4 adj (2A) . Then, det ( adj (X)) = ( det (X))^2 . Now, evaluate det (X) : det (4 adj (2A)) = 4^3 det ( adj (2A)) = 64 ( det (2A))^2 = 64 (2^3 det (A))^2 = 64 (8 det (A))^2 = 64 64 det (A)^2 = 2¹² det (A)^2 Substitute this back into the first equation: det ( adj (X)) = (2¹² det (A)^2)^2 = 2²⁴ det (A)^4 We are given that this equals 2³² : 2²⁴ d