If \( \mathrm{A}=\begin{bmatrix}1 & 0 & 0\\3 & 3 & 0\\5 & 2 & -1\end{bmatrix} \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( |\mathrm{adj\ A}| \) | I. | 81 |
| B. | \( |\mathrm{A}| \) | II. | \( -27 \) |
| C. | \( |\mathrm{A\ adj\ A}| \) | III. | 9 |
| D. | \( |\mathrm{adj\ (adj\ A)}| \) | IV. | \( -3 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-IV, C-II, D-I
A is lower triangular, so its determinant is the product of the diagonal entries.
\( \displaystyle |\mathrm{A}|=1\times 3\times(-1)=-3 \)
For a square matrix of order n = 3, \( |\mathrm{adj\,A}|=|\mathrm{A}|^{n-1} \), \( \mathrm{A}(\mathrm{adj\,A})=|\mathrm{A}|\,\mathrm{I} \) and \( |\mathrm{adj(adj\,A)}|=|\mathrm{A}|^{(n-1)^2} \).
(A) \( |\mathrm{adj\ A}| \)
This matches (III) 9.
\( \displaystyle |\mathrm{adj\,A}|=(-3)^2=9 \)
(B) \( |\mathrm{A}| \)
This matches (IV) −3.
(C) \( |\mathrm{A\ adj\ A}| \)
This matches (II) −27.
\( \displaystyle |\mathrm{A}(\mathrm{adj\,A})|=|-3\,\mathrm{I}| \)
\( \displaystyle =(-3)^3=-27 \)
(D) \( |\mathrm{adj\ (adj\ A)}| \)
This matches (I) 81.
\( \displaystyle |\mathrm{adj(adj\,A)}|=(-3)^{(3-1)^2} \)
\( \displaystyle =(-3)^4=81 \)
So the matching is A-III, B-IV, C-II, D-I.