A is a square matrix of order 3 × 3, then
Match List-I with List-II
| List-I | List-II | ||
| (A) | |adj A| | (I) | \( \displaystyle\frac{1}{|A|} \) |
| (B) | |A-1| | (II) | |A|4 |
| (C) | |A.adj A| | (III) | |A|3 |
| (D) | |adj(adj A)| | (IV) | |A|2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (I), (C) - (III), (D) - (II)
Here the order is n = 3. The results used are \( A(\mathrm{adj}\,A)=|A|\,I \), \( |\mathrm{adj}\,A|=|A|^{n-1} \) and \( |kM|=k^n|M| \).
(A) |adj A|
This matches (IV) |A|2.
\( \displaystyle |\mathrm{adj}\,A|=|A|^{n-1} \)
\( \displaystyle =|A|^2 \)
(B) |A-1|
This matches (I) \( \frac{1}{|A|} \).
\( \displaystyle |A|\,|A^{-1}|=|AA^{-1}|=|I|=1 \)
\( \displaystyle |A^{-1}|=\frac{1}{|A|} \)
(C) |A.adj A|
This matches (III) |A|3.
\( \displaystyle |A\,(\mathrm{adj}\,A)|=\big|\,|A|\,I\,\big| \)
\( \displaystyle =|A|^3\,|I| \)
\( \displaystyle =|A|^3 \)
(D) |adj(adj A)|
This matches (II) |A|4.
\( \displaystyle |\mathrm{adj}(\mathrm{adj}\,A)|=|\mathrm{adj}\,A|^{n-1} \)
\( \displaystyle =\left(|A|^2\right)^2 \)
\( \displaystyle =|A|^4 \)
So the matching is (A) - (IV), (B) - (I), (C) - (III), (D) - (II).
Tip: For order n, remember \( |\mathrm{adj}\,A|=|A|^{n-1} \) and \( |\mathrm{adj}(\mathrm{adj}\,A)|=|A|^{(n-1)^2} \); a scalar |A| taken out of a determinant of order n comes out as \( |A|^n \).