If A is a non-singular matrix of order 3, then
Match List-I with List-II
| List-I | List-II | ||
| Determinant of A | |adj(adj A)| = | ||
| (A) | |A| = 3 | (I) | 1 |
| (B) | |A| = 4 | (II) | 16 |
| (C) | |A| = 2 | (III) | 81 |
| (D) | |A| = 1 | (IV) | 256 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
For a non-singular matrix of order n, \( |\mathrm{adj}\,A|=|A|^{n-1} \). Applying this twice (adj A is also of order n):
\( \displaystyle |\mathrm{adj}(\mathrm{adj}\,A)|=|\mathrm{adj}\,A|^{n-1} \)
\( \displaystyle =|A|^{(n-1)^2} \)
With n = 3 this is \( |A|^4 \).
(A) |A| = 3
This matches (III) 81.
\( \displaystyle 3^4=81 \)
(B) |A| = 4
This matches (IV) 256.
\( \displaystyle 4^4=256 \)
(C) |A| = 2
This matches (II) 16.
\( \displaystyle 2^4=16 \)
(D) |A| = 1
This matches (I) 1.
\( \displaystyle 1^4=1 \)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).
Tip: Remember the exponent \( (n-1)^2 \), not \( 2(n-1) \): for n = 3 both happen to be 4, but for n = 4 they are 9 and 6.