Let \( A=[a_{ij}]_{2\times 2} \) such that \( a_{ij}=2i+j \) for all \( i \) and \( j \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( |A| \) | (I) | -1/2 |
| (B) | \( |A^{2}| \) | (II) | 16 |
| (C) | \( |2A^{2}| \) | (III) | -2 |
| (D) | \( \left|\displaystyle\frac{A}{2}\right| \) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
From \( a_{ij}=2i+j \):
\( A=\begin{bmatrix}3 & 4\\ 5 & 6\end{bmatrix} \)
The values use \( |A^{2}|=|A|^{2} \) and, for a matrix of order 2, \( |kA|=k^{2}|A| \).
(A) \( |A| \)
This matches (III).
\( |A|=18-20=-2 \)
(B) \( |A^{2}| \)
This matches (IV).
\( |A^{2}|=(-2)^{2}=4 \)
(C) \( |2A^{2}| \)
This matches (II).
\( |2A^{2}|=2^{2}\times 4=16 \)
(D) \( \left|\frac{A}{2}\right| \)
This matches (I).
\( \displaystyle \left|\frac{A}{2}\right|=\left(\frac{1}{2}\right)^{2}(-2)=-\frac{1}{2} \)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).