If A and B are invertible matrices of the same order, then,
Match List-I with List-II
| LIST-I | LIST-II |
| A. \( |A^T| \) | I. \( B^{-1}\ A^{-1} \) |
| B. \( A(adj\ A) \) | II. \( adj\ A \) |
| C. \( A^{-1}\ |A| \) | III. \( (adj\ A)A \) |
| D. \( (AB)^{-1} \) | IV. \( |A| \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-III, C-II, D-I
(A) \( |A^T| \)
This matches IV. |A|. A matrix and its transpose have the same determinant.
(B) \( A(adj\ A) \)
This matches III. (adj A)A. Both products equal \( |A|\,I \).
(C) \( A^{-1}\ |A| \)
This matches II. adj A. Since \( A^{-1}=\frac{1}{|A|}\mathrm{adj}\,A \), multiplying by |A| gives adj A.
(D) \( (AB)^{-1} \)
This matches I. \( B^{-1}A^{-1} \). This is the reversal law for inverses.
So the matching is A-IV, B-III, C-II, D-I.