# Construct a 4 × 3 matrix whose elements are

Question:

Construct a $4 \times 3$ matrix whose elements are

(i) $a_{i j}=2 i+\frac{i}{j}$

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

(iii) $a_{i j}=i$

Solution:

(i)

$a_{i j}=2 i+\frac{i}{j}$

Here,

$a_{11}=2(1)+\frac{1}{1}=\frac{2+1}{1}=\frac{3}{1}=3, a_{12}=2(1)+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}, a_{13}=2(1)+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$

$a_{21}=2(2)+\frac{2}{1}=\frac{4+2}{1}=\frac{6}{1}=6, a_{22}=2(2)+\frac{2}{2}=\frac{8+2}{2}=\frac{10}{2}=5, a_{23}=2(2)+\frac{2}{3}=\frac{12+2}{3}=\frac{14}{3}$

$a_{31}=2(3)+\frac{3}{1}=\frac{6+3}{1}=\frac{9}{1}=9, a_{32}=2(3)+\frac{3}{2}=\frac{12+3}{2}=\frac{15}{2}, a_{33}=2(3)+\frac{3}{3}=\frac{18+3}{3}=\frac{21}{3}=7$

$a_{41}=2(4)+\frac{4}{1}=\frac{8+4}{1}=\frac{12}{1}=12, a_{42}=2(4)+\frac{4}{2}=\frac{16+4}{2}=\frac{20}{2}=10$ and $a_{43}=2(4)+\frac{4}{3}=\frac{24+4}{3}=\frac{28}{3}$

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

(ii)

$a_{i j}=\frac{i-j}{i+j}$

Here,

$a_{11}=\frac{1-1}{1+1}=\frac{0}{2}=0, a_{12}=\frac{1-2}{1+2}=\frac{-1}{3}, a_{13}=\frac{1-3}{1+3}=\frac{-2}{4}=\frac{-1}{2}$

$a_{21}=\frac{2-1}{2+1}=\frac{1}{3}, a_{22}=\frac{2-2}{2-2}=\frac{0}{0}=0, a_{23}=\frac{2-3}{2+3}=\frac{-1}{5}$

$a_{31}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}, a_{32}=\frac{3-2}{3+2}=\frac{1}{5}, a_{33}=\frac{3-3}{3+3}=\frac{0}{6}=0$

$a_{41}=\frac{4-1}{4+1}=\frac{3}{5}, a_{42}=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}$ and $a_{43}=\frac{4-3}{4+3}=\frac{1}{7}$

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

(iii)

$a_{i j}=i$

Here,

$a_{11}=1, a_{12}=1, a_{13}=1$

$a_{21}=2, a_{22}=2, a_{23}=2$

$a_{31}=3, a_{32}=3, a_{33}=3$

$a_{41}=4, a_{42}=4$ and $a_{43}=4$

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