Match List-I with List-II
| List-I | List-II | ||
| Matrix A | \( |\mathrm{adj}\ A|^2 \) | ||
| (A) | \( \begin{bmatrix}3 & 7\\2 & 8\end{bmatrix} \) | (I) | 121 |
| (B) | \( \begin{bmatrix}5 & -1\\2 & 4\end{bmatrix} \) | (II) | 144 |
| (C) | \( \begin{bmatrix}3 & 6\\5 & 14\end{bmatrix} \) | (III) | 484 |
| (D) | \( \begin{bmatrix}7 & 5\\2 & 3\end{bmatrix} \) | (IV) | 100 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
For a square matrix of order n, \( |\mathrm{adj}\,A|=|A|^{n-1} \). For n = 2 this gives \( |\mathrm{adj}\,A|=|A| \), so \( |\mathrm{adj}\,A|^2=|A|^2 \).
(A) \( \begin{bmatrix}3 & 7\\2 & 8\end{bmatrix} \)
This matches (IV) 100.
\( \displaystyle |A|=24-14=10 \)
\( \displaystyle |A|^2=100 \)
(B) \( \begin{bmatrix}5 & -1\\2 & 4\end{bmatrix} \)
This matches (III) 484.
\( \displaystyle |A|=20+2=22 \)
\( \displaystyle |A|^2=484 \)
(C) \( \begin{bmatrix}3 & 6\\5 & 14\end{bmatrix} \)
This matches (II) 144.
\( \displaystyle |A|=42-30=12 \)
\( \displaystyle |A|^2=144 \)
(D) \( \begin{bmatrix}7 & 5\\2 & 3\end{bmatrix} \)
This matches (I) 121.
\( \displaystyle |A|=21-10=11 \)
\( \displaystyle |A|^2=121 \)
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).