Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | If A is an invertible square matrix then \( |\mathrm{A}^{-1}| \) = | I. | \( \displaystyle\frac{1}{|\mathrm{A}|} \) (adj A) |
| B. | For every square matrix (adj A) A = | II. | \( \displaystyle\frac{1}{|\mathrm{A}|} \) |
| C. | For a square matrix A, if \( |\mathrm{A}|\ne 0 \) , then \( \mathrm{A}^{-1} \) = | III. | \( \mathrm{B}^{-1}\mathrm{A}^{-1} \) |
| D. | If A and B are invertible matrices, then \( (\mathrm{AB})^{-1} \) = | IV. | \( |\mathrm{A}|.\mathrm{I} \), where I is the identity matrix of same order as A. |
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
(A) If A is an invertible square matrix then \( |\mathrm{A}^{-1}| \) =
This matches (II) \( \frac{1}{|\mathrm{A}|} \).
\( \displaystyle \mathrm{A}\mathrm{A}^{-1}=\mathrm{I} \)
\( \displaystyle \Rightarrow\ |\mathrm{A}|\,|\mathrm{A}^{-1}|=|\mathrm{I}|=1 \)
\( \displaystyle \Rightarrow\ |\mathrm{A}^{-1}|=\frac{1}{|\mathrm{A}|} \)
(B) For every square matrix (adj A) A =
This matches (IV) |A| I.
\( \displaystyle \mathrm{A}(\mathrm{adj\,A})=(\mathrm{adj\,A})\mathrm{A}=|\mathrm{A}|\,\mathrm{I} \)
(C) For a square matrix A, if |A| ≠ 0, then \( \mathrm{A}^{-1} \) =
This matches (I) \( \frac{1}{|\mathrm{A}|} \) (adj A).
Dividing \( (\mathrm{adj\,A})\mathrm{A}=|\mathrm{A}|\,\mathrm{I} \) by |A| ≠ 0 gives
\( \displaystyle \mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}\,\mathrm{adj\,A} \)
(D) If A and B are invertible matrices, then \( (\mathrm{AB})^{-1} \) =
This matches (III) \( \mathrm{B}^{-1}\mathrm{A}^{-1} \).
\( \displaystyle (\mathrm{AB})(\mathrm{B}^{-1}\mathrm{A}^{-1})=\mathrm{A}(\mathrm{B}\mathrm{B}^{-1})\mathrm{A}^{-1} \)
\( \displaystyle =\mathrm{A}\mathrm{A}^{-1}=\mathrm{I} \)
So the matching is A-II, B-IV, C-I, D-III.