Match List I with List II: Given that A and B are invertible matrices of size 3 × 3
| List I | List II | ||
| A. | \(|AB|\) | I. | \(\displaystyle\frac{1}{|A|}\) |
| B. | \(|\mathrm{Adj}\,A|\) | II. | \(|A|\,|B|\) |
| C. | \(|A^{-1}|\) | III. | \(B^{-1} \cdot A^{-1}\) |
| D. | \((AB)^{-1}\) | IV. | \(|A|^2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-IV, C-I, D-III
(A) \(|AB|\)
This matches (II) \(|A|\,|B|\). The determinant of a product is the product of the determinants.
(B) \(|\mathrm{Adj}\,A|\)
This matches (IV) \(|A|^2\). For order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\); here n = 3, so the power is 2.
(C) \(|A^{-1}|\)
This matches (I) \(\frac{1}{|A|}\).
\(|A|\,|A^{-1}| = |AA^{-1}| = |I| = 1\)
(D) \((AB)^{-1}\)
This matches (III) \(B^{-1} \cdot A^{-1}\). The inverse of a product reverses the order:
\((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I\)
So the matching is A-II, B-IV, C-I, D-III.