Let \( A=[a_{ij}] \) be a matrix of order \( 3\times 4 \) such that \( a_{ij}=\begin{cases}2i-3j, & i\ge j\\3i+j, & i\lt j\end{cases} \) then the sum of all the elements of the first two rows of this matrix is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 35
The first two rows have i = 1 and i = 2, with j = 1, 2, 3, 4. The rule is \( 2i-3j \) on and below the diagonal (i ≥ j) and \( 3i+j \) above it (i < j).
Row 1 (i = 1): only j = 1 has i ≥ j.
\( a_{11}=2-3=-1 \)
\( a_{12}=3+2=5 \)
\( a_{13}=3+3=6 \)
\( a_{14}=3+4=7 \)
\( \text{Row 1 sum}=-1+5+6+7=17 \)
Row 2 (i = 2): j = 1 and j = 2 have i ≥ j.
\( a_{21}=4-3=1 \)
\( a_{22}=4-6=-2 \)
\( a_{23}=6+3=9 \)
\( a_{24}=6+4=10 \)
\( \text{Row 2 sum}=1-2+9+10=18 \)
\( \text{Total}=17+18=35 \)