Let A and B be two square matrices of order 3 such that |A| = 3 and |B| = 2. Then |A^T adj(adj(2A))^-1 (adj(4B))(adj(AB))^-1 A^T| is equal to :
Step-by-Step Solution
Key Concept: Use properties of determinants: |kA| = k^n|A|, |adj(A)| = |A|^(n-1), |A^T| = |A|, |AB| = |A||B|, |A^-1| = 1/|A|.
Given |A|=3, |B|=2, n=3. The expression is |A^T| * |adj(adj(2A))^-1| * |adj(4B)| * |(adj(AB))^-1| * |A^T|. |A^T| = |A| = 3. |adj(adj(2A))^-1| = 1/|adj(adj(2A))| = 1/(|2A|^(n-1)^2) = 1/(2^3 * 3)^(3-1)^2 = 1/(24^2)^2 = 1/24^4. |adj(4B)| = |4B|^(3-1) = (4^3 * 2)^2 = 128^2. |(adj(AB))^-1| = 1/|adj(AB)| = 1/|AB|^(3-1) = 1/(|A||B|)^2 = 1/(3*2)^2 = 1/36. Combining these, the result simplifies to 64.
Correct Answer: 1