If A is a 3×3 matrix then:
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( |\mathrm{adj}A| \) | I. | \( |A|^{-1} \) |
| B. | \( |A.\mathrm{adj}A| \) | II. | \( |A| \) |
| C. | \( |A^{-1}| \) | III. | \( |A|^{2} \) |
| D. | \( |A^{-1}.\mathrm{adj}A| \) | IV. | \( |A|^{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-I, D-II
Items (C) and (D) contain \( A^{-1} \), so A is taken to be non-singular (\( |A|\ne 0 \)). For a square matrix of order n = 3, the standard results are
\( A(\mathrm{adj}A)=|A|\,I \)
\( |\mathrm{adj}A|=|A|^{n-1} \)
\( |kI|=k^{n} \)
\( |PQ|=|P|\,|Q| \)
(A) \( |\mathrm{adj}A| \)
This matches (III) \( |A|^{2} \).
\( |\mathrm{adj}A|=|A|^{3-1}=|A|^{2} \)
(B) \( |A.\mathrm{adj}A| \)
This matches (IV) \( |A|^{3} \).
\( |A.\mathrm{adj}A|=\Big|\,|A|\,I\,\Big| \)
\( =|A|^{3}|I|=|A|^{3} \)
(C) \( |A^{-1}| \)
This matches (I) \( |A|^{-1} \).
\( |A|\,|A^{-1}|=|AA^{-1}|=|I|=1 \)
\( |A^{-1}|=|A|^{-1} \)
(D) \( |A^{-1}.\mathrm{adj}A| \)
This matches (II) \( |A| \).
\( |A^{-1}.\mathrm{adj}A|=|A^{-1}|\,|\mathrm{adj}A| \)
\( =|A|^{-1}\cdot|A|^{2}=|A| \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Tip: Remember that \( |kI|=k^{n} \), not k: \( |A.\mathrm{adj}A| \) is \( |A|^{3} \) for a 3 × 3 matrix, although \( A.\mathrm{adj}A \) itself is \( |A|I \).