Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Matrix \( A \) | \( |\mathrm{adj}\ A| \) | ||
| A. | \( A=\begin{bmatrix}2 & 4\\1 & 3\end{bmatrix} \) | I. | 1 |
| B. | \( A=\begin{bmatrix}5 & 2\\7 & 4\end{bmatrix} \) | II. | 7 |
| C. | \( A=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \) | III. | 2 |
| D. | \( A=\begin{bmatrix}6 & 1\\5 & 2\end{bmatrix} \) | IV. | 6 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A - III, B - IV, C - I, D - II
For a square matrix of order n, \( |\mathrm{adj\,A}|=|\mathrm{A}|^{n-1} \). Here n = 2, so \( |\mathrm{adj\,A}|=|\mathrm{A}| \).
(A) \( \begin{bmatrix}2 & 4\\1 & 3\end{bmatrix} \)
This matches (III) 2.
\( |A|=6-4=2 \)
(B) \( \begin{bmatrix}5 & 2\\7 & 4\end{bmatrix} \)
This matches (IV) 6.
\( |A|=20-14=6 \)
(C) \( \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \)
This matches (I) 1.
\( |A|=1-0=1 \)
(D) \( \begin{bmatrix}6 & 1\\5 & 2\end{bmatrix} \)
This matches (II) 7.
\( |A|=12-5=7 \)
So the matching is A-III, B-IV, C-I, D-II.