Question
Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by: $a_{ij} = i + j$
✓
Answer
Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 1 + 1 = 2, a_{12} = 1 + 2 = 3, a_{13} = 1 + 3 = 4, a_{14} = 1 + 4 = 5$
$a_{21} = 2 + 1 = 3, a_{22} = 2 + 2 = 4, a_{23} = 2 + 3 = 5, a_{24} = 2 + 4 = 6$
$a_{31} = 3 + 1 = 4, a_{32} = 3 + 2 = 5, a_{33} = 3 + 3 = 6, a_{34} = 3 + 4 = 7$
Using equation $(i)$
$\text{A}=\begin{bmatrix}2&3&4&5\\3&4&5&6\\4&5&6&7\end{bmatrix}$
$a_{11} = 1 + 1 = 2, a_{12} = 1 + 2 = 3, a_{13} = 1 + 3 = 4, a_{14} = 1 + 4 = 5$
$a_{21} = 2 + 1 = 3, a_{22} = 2 + 2 = 4, a_{23} = 2 + 3 = 5, a_{24} = 2 + 4 = 6$
$a_{31} = 3 + 1 = 4, a_{32} = 3 + 2 = 5, a_{33} = 3 + 3 = 6, a_{34} = 3 + 4 = 7$
Using equation $(i)$
$\text{A}=\begin{bmatrix}2&3&4&5\\3&4&5&6\\4&5&6&7\end{bmatrix}$
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