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Let A=[aij]A=[a_{ij}] be a matrix of order 3×43 \times 4 such that aij={2i−3j,i≥j3i+j,i<ja_{ij}=\begin{cases}2i-3j, & i \ge j \\ 3i+j, & i<j\end{cases} then the sum of all the elements of the first two rows of this matrix is:

Solution

✅ Correct Option: 4

Row 1 (i=1i=1): a11=2−3=−1a_{11}=2-3=-1; a12=3+2=5a_{12}=3+2=5; a13=3+3=6a_{13}=3+3=6; a14=3+4=7a_{14}=3+4=7. Sum =17=17.

Row 2 (i=2i=2): a21=4−3=1a_{21}=4-3=1; a22=4−6=−2a_{22}=4-6=-2; a23=6+3=9a_{23}=6+3=9; a24=6+4=10a_{24}=6+4=10. Sum =18=18.

Total =17+18=35= 17+18 = 35.

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