If A and B are non-singular square matrices of order n, then
[Where I is an identity matrix of order n]
Match List-I with List-II
| List-I | List-II |
| (A) \( |\ (A^T)^{-1}| \) | (I) \( adj\ A \) |
| (B) \( adj(AB) \) | (II) \( \displaystyle\frac{1}{|A|} \) |
| (C) \( A\ (adj\ A) \) | (III) \( |A|\mathrm{I_n} \) |
| (D) \( A^{-1}\ |A| \) | (IV) \( adj\ B\ .\ adj\ A \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (IV), (C) - (III), (D) - (I)
(A) \( |(A^T)^{-1}| \)
This matches (II) \( \displaystyle\frac{1}{|A|} \), since \( |A^T|=|A| \):
\( \displaystyle |(A^T)^{-1}|=\frac{1}{|A^T|} \)
\( \displaystyle =\frac{1}{|A|} \)
(B) \( \mathrm{adj}(AB) \)
This matches (IV) \( \mathrm{adj}\,B\cdot\mathrm{adj}\,A \). This is the reversal law for adjoints.
(C) \( A\,(\mathrm{adj}\,A) \)
This matches (III) \( |A|\,I_n \). This is the basic property of the adjoint.
(D) \( A^{-1}\,|A| \)
This matches (I) \( \mathrm{adj}\,A \), since \( A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \).
So the matching is (A) - (II), (B) - (IV), (C) - (III), (D) - (I).