Let A be a square matrix of order 3 with \(|A| = -2\). Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Expression | Value | ||
| (A) | \(|\mathrm{adj}(A^{T})|\) | (I) | \(-2\) |
| (B) | \(|\mathrm{adj}(2A)|\) | (II) | \(\displaystyle \frac{1}{4}\) |
| (C) | \(|(\mathrm{adj}\,A)^{-1}|\) | (III) | 256 |
| (D) | \(|A^{-1}\,\mathrm{adj}\,A|\) | (IV) | 4 |
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)-(II), (D)-(I)
For any square matrix M of order 3, \(|\mathrm{adj}\,M| = |M|^{2}\). Here:
\(|\mathrm{adj}\,A| = (-2)^{2} = 4\)
(A) \(|\mathrm{adj}(A^{T})|\)
This matches (IV) 4.
Apply the rule to M = \(A^{T}\), whose determinant is \(|A^{T}| = |A| = -2\):
\(|\mathrm{adj}(A^{T})| = (-2)^{2} = 4\)
(B) \(|\mathrm{adj}(2A)|\)
This matches (III) 256.
\(|2A| = 2^{3} \times (-2) = -16\)
\(|\mathrm{adj}(2A)| = (-16)^{2} = 256\)
(C) \(|(\mathrm{adj}\,A)^{-1}|\)
This matches (II) \(\displaystyle \frac{1}{4}\).
\(\displaystyle |(\mathrm{adj}\,A)^{-1}| = \frac{1}{|\mathrm{adj}\,A|} = \frac{1}{4}\)
(D) \(|A^{-1}\,\mathrm{adj}\,A|\)
This matches (I) \(-2\).
\(\displaystyle |A^{-1}\,\mathrm{adj}\,A| = |A^{-1}|\,|\mathrm{adj}\,A| = \frac{1}{-2} \times 4 = -2\)