Match the LIST-I with LIST-II
| LIST-I Matrix A = | LIST-II \( |adjA|= \) | ||
| A. | \( \begin{bmatrix}2 & 1\\1 & 3\end{bmatrix} \) | I. | 3 |
| B. | \( \begin{bmatrix}1 & 0\\3 & 2\end{bmatrix} \) | II. | 2 |
| C. | \( \begin{bmatrix}3 & 0\\0 & 1\end{bmatrix} \) | III. | 5 |
| D. | \( \begin{bmatrix}0 & -1\\4 & 0\end{bmatrix} \) | IV. | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-II, C-I, D-IV
For a square matrix of order n, \( |\mathrm{adj}\,A|=|A|^{n-1} \). For a 2 × 2 matrix this is just |A|.
(A) \( \begin{bmatrix}2 & 1\\1 & 3\end{bmatrix} \)
This matches III. 5.
\( \displaystyle |A|=6-1=5 \)
(B) \( \begin{bmatrix}1 & 0\\3 & 2\end{bmatrix} \)
This matches II. 2.
\( \displaystyle |A|=2-0=2 \)
(C) \( \begin{bmatrix}3 & 0\\0 & 1\end{bmatrix} \)
This matches I. 3.
\( \displaystyle |A|=3-0=3 \)
(D) \( \begin{bmatrix}0 & -1\\4 & 0\end{bmatrix} \)
This matches IV. 4.
\( \displaystyle |A|=0-(-4)=4 \)
So the matching is A-III, B-II, C-I, D-IV.