Question

If A and B are non-singular square matrices of order n, match List-I with List-II:

• A. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
• B. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
• C. (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
• D. (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

Hint:

Remember: \( adj(AB) = adj(B)adj(A) \). Note the reversal of order, similar to the property of matrix inversion \((AB)^{-1} = B^{-1}A^{-1}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We evaluate each matrix identity using fundamental properties of determinants, adjoints, and inverses. 

Step 2: Detailed Explanation: 

• (A) \( |(A^T)^{-1| \):} Since \( |A^T| = |A| \) and \( |A^{-1}| = 1/|A| \), we have \( |(A^T)^{-1}| = 1/|A^T| = 1/|A| \). This matches (II). 

• (B) \( adj(AB) \): By the property of adjoints of products, \( adj(AB) = adj(B) \cdot adj(A) \). This matches (IV). 

• (C) \( A(adj A) \): This is the standard identity \( A(adj A) = |A|I_n \). This matches (III). 

• (D) \( A^{-1} |A| \): Starting from \( A(adj A) = |A|I \), multiplying both sides by \( A^{-1} \) yields \( adj A = A^{-1} |A| \). This matches (I). 

Step 3: Final Answer: 
The correct matching is (A)-(II), (B)-(IV), (C)-(III), (D)-(I), which corresponds to option (C).