Question
Construct a 4 × 3 matrix whose element are:
$\text{a}_\text{ij}=2\text{i}+\frac{\text{i}}{\text{j}}$

Answer

Here,
$\text{a}_{11}=2(1)+\frac{1}{1}=\frac{2+1}{1}=\frac{3}{1}=3,$ $\text{a}_{12}=2(1)+\frac{1}{3}=\frac{4+1}{2}=\frac{5}{2},$ $\text{a}_{13}=2(1)+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$
$\text{a}_{21}=2(2)+\frac{2}{1}=\frac{4+2}{1}=\frac{6}{1}=6,$ $\text{a}_{22}=2(2)+\frac{2}{2}=\frac{8+2}{2}=\frac{10}{2}=5,$ $\text{a}_{23}=2(2)+\frac{2}{3}=\frac{12+2}{3}=\frac{14}{3}$
$\text{a}_{31}=2(3)+\frac{3}{1}=\frac{6+3}{1}=\frac{9}{1}=9,$ $\text{a}_{32}=2(3)+\frac{3}{2}=\frac{12+3}{2}=\frac{15}{2},$ $\text{a}_{33}=2(3)+\frac{3}{5}=\frac{18+3}{3}=\frac{21}{3}=7$
$\text{a}_{41}=2(4)+\frac{4}{1}=\frac{8+4}{1}=\frac{12}{1}=12,$ $\text{a}_{42}=2(4)+\frac{4}{2}=\frac{16+4}{2}=\frac{20}{2}=10$ and $\text{a}_{43}=2(4)+\frac{4}{3}=\frac{24+4}{3}=\frac{28}{3}$
So, the required matrix is $\begin{bmatrix}3&\frac{5}{2}&\frac{7}{3}\\6&5&\frac{14}{3}\\9&\frac{15}{2}&7\\12&10&\frac{28}{3}\end{bmatrix}.$

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