If A and B are invertible matrices of the same order, then
Match List-I with List-II
| List-I | List-II | ||
| Determinant | Equal to | ||
| (A) | |A-1B| | (I) | \( \displaystyle\frac{1}{|A||B|} \) |
| (B) | |AB-1| | (II) | \( \displaystyle\frac{|B|}{|A|} \) |
| (C) | |A-1 B-1| | (III) | \( |A||B| \) |
| (D) | |AB| | (IV) | \( \displaystyle\frac{|A|}{|B|} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
The results used are \( |XY|=|X|\,|Y| \) and \( |X^{-1}|=\frac{1}{|X|} \).
(A) |A-1B|
This matches (II) \( \frac{|B|}{|A|} \).
\( \displaystyle |A^{-1}B|=|A^{-1}|\,|B| \)
\( \displaystyle =\frac{|B|}{|A|} \)
(B) |AB-1|
This matches (IV) \( \frac{|A|}{|B|} \).
\( \displaystyle |AB^{-1}|=|A|\,|B^{-1}| \)
\( \displaystyle =\frac{|A|}{|B|} \)
(C) |A-1 B-1|
This matches (I) \( \frac{1}{|A||B|} \).
\( \displaystyle |A^{-1}B^{-1}|=\frac{1}{|A|}\cdot\frac{1}{|B|} \)
\( \displaystyle =\frac{1}{|A||B|} \)
(D) |AB|
This matches (III) \( |A||B| \).
\( \displaystyle |AB|=|A|\,|B| \)
So the matching is (A) - (II), (B) - (IV), (C) - (I), (D) - (III).