If \( A=\left[a_{ij}\right]_{3\times 3} \) and \( a_{ij}=2i-j \), then
Match List-I with List-II
| List-I | List-II |
| (A) \( a_{21}+a_{32} \) | (I) 0 |
| (B) \( |A| \) | (II) 19 |
| (C) Number of the elements in A | (III) 7 |
| (D) \( a_{22}\cdot a_{31}+a_{21}\cdot a_{33} \) | (IV) 9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Putting i, j = 1, 2, 3 in \( a_{ij}=2i-j \):
\( \displaystyle A=\begin{bmatrix}1 & 0 & -1\\3 & 2 & 1\\5 & 4 & 3\end{bmatrix} \)
(A) \( a_{21}+a_{32} \)
This matches (III) 7.
\( \displaystyle a_{21}+a_{32}=3+4=7 \)
(B) \( |A| \)
This matches (I) 0. Subtracting row 1 from row 2, and row 2 from row 3, gives two equal rows (2, 2, 2), so
\( \displaystyle |A|=0 \)
(C) Number of the elements in A
This matches (IV) 9. A 3 × 3 matrix has 3 × 3 = 9 elements.
(D) \( a_{22}\cdot a_{31}+a_{21}\cdot a_{33} \)
This matches (II) 19.
\( \displaystyle a_{22}\cdot a_{31}+a_{21}\cdot a_{33}=2\times 5+3\times 3=19 \)
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).