If matrix \( A=\begin{bmatrix}1 & 4 & 5\\3 & 2 & 6\\0 & 1 & 0\end{bmatrix} \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( |A|= \) | (I) | \( \displaystyle\frac{1}{9} \) |
| (B) | \( |A^{-1}|= \) | (II) | 81 |
| (C) | \( |\mathrm{adj}\,A|= \) | (III) | 9 |
| (D) | \( |\mathrm{adj}\,(\mathrm{adj}\,A)|= \) | (IV) | 6561 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
(A) |A|
This matches (III) 9.
Expanding along R3, which has two zeros (only the middle entry 1 contributes, with sign \( (-1)^{3+2}=-1 \)):
\( \displaystyle |A|=-1\cdot\begin{vmatrix}1 & 5\\3 & 6\end{vmatrix} \)
\( \displaystyle =-(6-15)=9 \)
(B) |A−1|
This matches (I) \( \frac{1}{9} \).
\( \displaystyle |A^{-1}|=\frac{1}{|A|}=\frac{1}{9} \)
(C) |adj A|
This matches (II) 81.
\( \displaystyle |\mathrm{adj}\,A|=|A|^{3-1}=9^2=81 \)
(D) |adj (adj A)|
This matches (IV) 6561.
\( \displaystyle |\mathrm{adj}(\mathrm{adj}\,A)|=|A|^{(3-1)^2}=9^4=6561 \)
So the matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV).