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If A=[aij]3×3A=[a_{ij}]_{3\times3} and aij=2i−ja_{ij}=2i-j, then

Match List-I with List-II

List-IList-II
(A) a21+a32a_{21}+a_{32}(I) 0
(B) ∣A∣\lvert A\rvert(II) 19
(C) Number of the elements in A(III) 7
(D) a22.a31+a21.a33a_{22}.a_{31}+a_{21}.a_{33}(IV) 9

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

The rows are (1,0,−1)(1,0,-1), (3,2,1)(3,2,1), (5,4,3)(5,4,3). So a21+a32=3+4=7a_{21}+a_{32}=3+4=7; the determinant is 0 because the third row equals twice the second minus the first; there are 3×3=93\times3=9 elements; and a22a31+a21a33=2(5)+3(3)=19a_{22}a_{31}+a_{21}a_{33}=2(5)+3(3)=19.

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