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If A=[aij]3×3A = [a_{ij}]_{3\times3} is a square matrix, where aij=i−j+3a_{ij} = i - j + 3, then the value of a21+a31−a23a_{21} + a_{31} - a_{23} is

Solution

✅ Correct Option: 3

Substituting into aij=i−j+3a_{ij}=i-j+3: a21=2−1+3=4a_{21}=2-1+3=4, a31=3−1+3=5a_{31}=3-1+3=5 and a23=2−3+3=2a_{23}=2-3+3=2. Hence a21+a31−a23=4+5−2=7a_{21}+a_{31}-a_{23}=4+5-2=7.

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