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) $|A^T|=|A|$ (IV)
(B) $A(adj\,A)=(adj\,A)A$ (III)
(C) $A^{-1}|A|=(adj\,A)$
as $A\,adj\,A=I|A|$
pre multiplying with $A^{-1}$
$adj\,A=A^{-1}|A|$ (II)
(D) $(AB)^{-1}=B^{-1}A^{-1}$ (I)