Question

Construct a 4 × 3 matrix whose elements are
(i) $a_{ij}=2i+\frac{i}{j}$

(ii) $a_{ij}=\frac{i-j}{i+j}$

(iii) a_ij = i

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right) \\ {a}_{ij}=2i+\frac{i}{j} \end{gathered}$$

Here,
$$\begin{gathered} a_{11}=2\left(1\right)+\frac{1}{1}=\frac{2+1}{1}=\frac{3}{1}=3,a_{12}=2\left(1\right)+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2},a_{13}=2\left(1\right)+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3} \\ {a}_{21}=2\left(2\right)+\frac{2}{1}=\frac{4+2}{1}=\frac{6}{1}=6,a_{22}=2\left(2\right)+\frac{2}{2}=\frac{8+2}{2}=\frac{10}{2}=5,a_{23}=2\left(2\right)+\frac{2}{3}=\frac{12+2}{3}=\frac{14}{3} \\ {a}_{31}=2\left(3\right)+\frac{3}{1}=\frac{6+3}{1}=\frac{9}{1}=9,a_{32}=2\left(3\right)+\frac{3}{2}=\frac{12+3}{2}=\frac{15}{2},a_{33}=2\left(3\right)+\frac{3}{3}=\frac{18+3}{3}=\frac{21}{3}=7 \\ {a}_{41}=2\left(4\right)+\frac{4}{1}=\frac{8+4}{1}=\frac{12}{1}=12,a_{42}=2\left(4\right)+\frac{4}{2}=\frac{16+4}{2}=\frac{20}{2}=10\mathrm{and}{a}_{43}=2\left(4\right)+\frac{4}{3}=\frac{24+4}{3}=\frac{28}{3} \end{gathered}$$


So, the required matrix is $\left[\begin{array}{ccc}3& \frac{5}{2}& \frac{7}{3}\\ 6& 5& \frac{14}{3}\\ 9& \frac{15}{2}& 7\\ 12& 10& \frac{28}{3}\end{array}\right]$.

$$\begin{gathered} \left(\mathrm{ii}\right) \\ {a}_{ij}=\frac{i-j}{i+j} \end{gathered}$$

Here,
$$\begin{gathered} a_{11}=\frac{1-1}{1+1}=\frac{0}{2}=0,a_{12}=\frac{1-2}{1+2}=\frac{-1}{3},a_{13}=\frac{1-3}{1+3}=\frac{-2}{4}=\frac{-1}{2} \\ {a}_{21}=\frac{2-1}{2+1}=\frac{1}{3},a_{22}=\frac{2-2}{2-2}=\frac{0}{0}=0,a_{23}=\frac{2-3}{2+3}=\frac{-1}{5} \\ {a}_{31}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2},a_{32}=\frac{3-2}{3+2}=\frac{1}{5},a_{33}=\frac{3-3}{3+3}=\frac{0}{6}=0 \\ {a}_{41}=\frac{4-1}{4+1}=\frac{3}{5},a_{42}=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\mathrm{and}{a}_{43}=\frac{4-3}{4+3}=\frac{1}{7} \end{gathered}$$


So, the required matrix is $\left[\begin{array}{ccc}0& \frac{-1}{3}& \frac{-1}{2}\\ \frac{1}{3}& 0& \frac{-1}{5}\\ \frac{1}{2}& \frac{1}{5}& 0\\ \frac{3}{5}& \frac{1}{3}& \frac{1}{7}\end{array}\right]$.

$$\begin{gathered} \left(\mathrm{iii}\right) \\ {a}_{ij}=i \end{gathered}$$

Here,
$$\begin{gathered} a_{11}=1,a_{12}=1,a_{13}=1 \\ {a}_{21}=2,a_{22}=2,a_{23}=2 \\ {a}_{31}=3,a_{32}=3,a_{33}=3 \\ {a}_{41}=4,a_{42}=4\mathrm{and}{a}_{43}=4 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{ccc}1& 1& 1\\ 2& 2& 2\\ 3& 3& 3\\ 4& 4& 4\end{array}\right]$.