Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | If A is a square matrix of order 3 such that \( |A|=3 \), then \( |\mathrm{adj}\ (\mathrm{adj}\ A)|= \) | I. | 16 |
| B. | If A is a non-singular matrix and \( |A|=\displaystyle\frac{1}{16} \), then \( |A^{-1}|= \) | II. | 512 |
| C. | If A is a square matrix of order 4 such that \( |A|=2 \), then \( |\mathrm{adj}\ (\mathrm{adj}\ A)|= \) | III. | 9 |
| D. | If A is a square matrix of order 2 such that \( |A|=9 \), then \( |\mathrm{adj}\ (\mathrm{adj}\ A)|= \) | IV. | 81 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A - IV, B - I, C - II, D - III
For a square matrix of order n, \( |\mathrm{adj\,A}|=|\mathrm{A}|^{n-1} \), so
\( |\mathrm{adj(adj\,A)}|=|\mathrm{adj\,A}|^{n-1} \)
\( |\mathrm{adj(adj\,A)}|=|\mathrm{A}|^{(n-1)^2} \)
and \( |\mathrm{A}^{-1}|=\frac{1}{|\mathrm{A}|} \).
(A) n = 3, |A| = 3
This matches (IV) 81.
\( 3^{(3-1)^2}=3^4=81 \)
(B) |A| = 1/16
This matches (I) 16.
\( \displaystyle |\mathrm{A}^{-1}|=\frac{1}{1/16}=16 \)
(C) n = 4, |A| = 2
This matches (II) 512.
\( 2^{(4-1)^2}=2^9=512 \)
(D) n = 2, |A| = 9
This matches (III) 9.
\( 9^{(2-1)^2}=9^1=9 \)
So the matching is A-IV, B-I, C-II, D-III.
Tip: For a 2 × 2 matrix, adj(adj A) = A itself, so \( |\mathrm{adj(adj\,A)}|=|\mathrm{A}| \); this settles (D) at a glance.