Let \( A=\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} \) and \( B=\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix} \) then
Match List-I with List-II
| List-I | List-II | ||
| Determinants | Value | ||
| (A) | \( |A^2| \) | (I) | 28 |
| (B) | \( |AB^T| \) | (II) | 7 |
| (C) | \( |\mathrm{adj}\,A| \) | (III) | 16 |
| (D) | \( |2B| \) | (IV) | 49 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
\( \displaystyle |A|=7-0=7 \)
\( \displaystyle |B|=0-(-4)=4 \)
(A) \( |A^2| \)
This matches (IV) 49.
\( \displaystyle |A^2|=|A|^2=49 \)
(B) \( |AB^T| \)
This matches (I) 28.
\( \displaystyle |AB^T|=|A|\,|B^T| \)
\( \displaystyle =|A|\,|B| \)
\( \displaystyle =7\times 4=28 \)
(C) \( |\mathrm{adj}\,A| \)
This matches (II) 7.
For a matrix of order n, \( |\mathrm{adj}\,A|=|A|^{n-1} \); here n = 2:
\( \displaystyle |\mathrm{adj}\,A|=7^{1}=7 \)
(D) \( |2B| \)
This matches (III) 16.
For a matrix of order 2, \( |kB|=k^2|B| \):
\( \displaystyle |2B|=2^2\times 4=16 \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).