If A is a square matrix of order \( n \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II |
| A. \( |A\,(adj\ A)| \) | I. \( |A|^{n-1} \) |
| B. \( adj\ (adj\ A) \) | II. \( |A|\mathrm{I} \), where I is the identity matrix of order \( n \) |
| C. \( |adj\ A| \) | III. \( |A|^{n-2}\ A \) |
| D. \( A(adj\ A) \) | IV. \( |A|^{n} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-IV, B-III, C-I, D-II
All four follow from the basic result \( A\,(\mathrm{adj}\,A)=|A|\,I \).
(A) \( |A\,(adj\ A)| \)
This matches IV. |A|n. The matrix |A| I has n diagonal entries equal to |A|:
\( \displaystyle |A\,(\mathrm{adj}\,A)|=\big||A|\,I\big| \)
\( \displaystyle =|A|^n \)
(B) \( adj\ (adj\ A) \)
This matches III. |A|n−2 A. Applying the basic result to adj A and using \( |\mathrm{adj}\,A|=|A|^{n-1} \):
\( \displaystyle \mathrm{adj}(\mathrm{adj}\,A)=|A|^{n-2}A \)
(C) \( |adj\ A| \)
This matches I. |A|n−1. Taking determinants in \( A\,(\mathrm{adj}\,A)=|A|\,I \):
\( \displaystyle |A|\,|\mathrm{adj}\,A|=|A|^n \)
\( \displaystyle |\mathrm{adj}\,A|=|A|^{n-1} \)
(D) \( A(adj\ A) \)
This matches II. |A| I. This is the basic result itself.
So the matching is A-IV, B-III, C-I, D-II.