Let A be a square matrix of order 3 such that \(|A| = 3\).
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|\mathrm{adj}\,A|\) | (I) | \(-3\) |
| (B) | \(|2A|\) | (II) | \(\displaystyle\frac{1}{3}\) |
| (C) | \(|A^{-1}|\) | (III) | 9 |
| (D) | \(|-A|\) | (IV) | 24 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
For a square matrix A of order n = 3 with \(|A| = 3\):
(A) \(|\mathrm{adj}\,A|\)
This matches (III) 9.
Using \(|\mathrm{adj}\,A| = |A|^{n-1}\):
\(|\mathrm{adj}\,A| = 3^{2} = 9\)
(B) \(|2A|\)
This matches (IV) 24.
Each of the 3 rows is multiplied by 2, so \(|kA| = k^{n}|A|\):
\(|2A| = 2^{3} \times 3 = 24\)
(C) \(|A^{-1}|\)
This matches (II) \(\frac{1}{3}\).
Since \(AA^{-1} = I\), \(|A|\,|A^{-1}| = 1\):
\( \displaystyle |A^{-1}| = \frac{1}{|A|} = \frac{1}{3} \)
(D) \(|-A|\)
This matches (I) \(-3\).
Here k = −1 and the order is odd:
\(|-A| = (-1)^{3} \times 3 = -3\)