For any non-zero matrix \( A=[a_{ij}]_{3\times 3} \) and \( |A|\ne 0 \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| (A) | \( |\mathrm{adj}\ A| \) | (I) | \( \displaystyle\frac{1}{|A|}(\mathrm{adj}\,A) \) |
| (B) | \( (\mathrm{adj}\ A)A \) | (II) | \( |A|^2 \) |
| (C) | \( A^{-1} \) | (III) | \( \displaystyle\frac{1}{|A|} \) |
| (D) | \( |A^{-1}| \) | (IV) | \( |A|\mathrm{I} \) |
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)
For a square matrix A of order n with \( |A|\ne 0 \), the standard results on the adjoint are
\( \displaystyle A(\mathrm{adj}\,A)=(\mathrm{adj}\,A)A=|A|\,\mathrm{I} \)
\( \displaystyle |\mathrm{adj}\,A|=|A|^{n-1} \)
\( \displaystyle |A^{-1}|=\frac{1}{|A|} \)
Here n = 3.
(A) \( |\mathrm{adj}\ A| \)
This matches (II) \( |A|^2 \).
\( \displaystyle |\mathrm{adj}\,A|=|A|^{3-1} \)
\( \displaystyle =|A|^2 \)
(B) \( (\mathrm{adj}\ A)A \)
This matches (IV) \( |A|\mathrm{I} \), since \( (\mathrm{adj}\,A)A=|A|\,\mathrm{I} \) for every square matrix.
(C) \( A^{-1} \)
This matches (I) \( \frac{1}{|A|}(\mathrm{adj}\,A) \).
Dividing \( (\mathrm{adj}\,A)A=|A|\,\mathrm{I} \) by \( |A| \) gives
\( \displaystyle A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A \)
(D) \( |A^{-1}| \)
This matches (III) \( \frac{1}{|A|} \).
\( \displaystyle |A|\,|A^{-1}|=|AA^{-1}| \)
\( \displaystyle =|\mathrm{I}|=1 \)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).