If \( A=[a_{ij}]_{3\times 3} \) , where \( a_{ij}=\begin{cases}i+j, & i\le j\\ i-j, & i\gt j\end{cases} \) then
Match List-I with List-II
| List-I | List-II |
| (A) \( |A| \) | (I) 176 |
| (B) \( |2A| \) | (II) 10648 |
| (C) \( |\mathrm{adj}\,A| \) | (III) 22 |
| (D) \( |A(\mathrm{adj}\,A)| \) | (IV) 484 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Above and on the diagonal \( a_{ij}=i+j \); below it \( a_{ij}=i-j \):
\( \displaystyle A=\begin{bmatrix}2 & 3 & 4\\1 & 4 & 5\\2 & 1 & 6\end{bmatrix} \)
(A) \( |A| \)
This matches (III) 22.
\( |A|=2(24-5)-3(6-10)+4(1-8) \)
\( =38+12-28 \)
\( =22 \)
(B) \( |2A| \)
This matches (I) 176. For a matrix of order 3, \( |kA|=k^3|A| \):
\( |2A|=2^3\times 22=176 \)
(C) \( |\mathrm{adj}\,A| \)
This matches (IV) 484. For order 3, \( |\mathrm{adj}\,A|=|A|^{2} \):
\( |\mathrm{adj}\,A|=22^2=484 \)
(D) \( |A(\mathrm{adj}\,A)| \)
This matches (II) 10648. Since \( A\,(\mathrm{adj}\,A)=|A|\,I \):
\( |A(\mathrm{adj}\,A)|=|A|^3 \)
\( =22^3=10648 \)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).