#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.1 Question 7 Subquestion (i) Maths Textbook Solution.

$A= \begin{bmatrix} 3 &\frac{5}{2} &\frac{7}{3} \\ \\ 6 &5 & \frac{14}{3}\\ \\ 9 & \frac{15}{2} &7 \\ \\ 12 & 10 & \frac{28}{3} \end{bmatrix}$

Given:   $a_{ij}= 2i+\frac{i}{j}$

Here we have to construct $4\times 3$  matrix according to  $a_{ij}= 2i+\frac{i}{j}$

Hint: First we will find all the elements of matrix according to $2i+\frac{i}{j}$

Solution: Here   $a_{ij}= 2i+\frac{i}{j}$

Let   $A= \left [ a_{ij} \right ]_{4\times 3}$

So, The elements in a $4\times 3$  matrix are $a_{11},a_{21},a_{31},a_{41},a_{12},a_{22},a_{32},a_{42},a_{13},a_{23},a_{33},a_{43}$

$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \\ a_{41} & a_{42} & a_{43} \end{bmatrix}_{4\times 3}$

$\! \! \! \! \! \! \! \! \! a_{11}= 2\times 1+\frac{1}{1}= 2+1= 3\\\\a_{12}= 2\times 1+\frac{1}{2}= 2+\frac{1}{2}= \frac{5}{2}\\\\a_{13}= 2\times 1+\frac{1}{3}= 2+\frac{1}{3}= \frac{7}{2}$                  $\! \! \! \! \! \! \! \! \! a_{21}= 2\times 2+\frac{2}{1}= 4+2= 6\\\\a_{22}= 2\times 2+\frac{2}{2}= 4+1= 5\\\\a_{23}= 2\times 2+\frac{2}{3}= 4+\frac{2}{3}= \frac{14}{3}$

$\! \! \! \! \! \! \! \! \! a_{31}= 2\times 3+\frac{3}{1}= 6+3= 9\\\\a_{32}= 2\times 3+\frac{3}{2}= 6+\frac{3}{2}= \frac{15}{2}\\\\a_{33}= 2\times 3+\frac{3}{3}= 6+1= 7$                $\! \! \! \! \! \! \! \! \! a_{41}= 2\times 4+\frac{4}{1}= 8+4= 12\\\\a_{42}= 2\times 4+\frac{4}{2}= 8+2= 10\\\\a_{43}= 2\times 4+\frac{4}{3}= 8+\frac{4}{3}= \frac{28}{3}$

Substituting these values in Matrix $A$ , we get

$A= \begin{bmatrix} 3 &\frac{5}{2} &\frac{7}{3} \\ \\ 6 &5 & \frac{14}{3}\\ \\ 9 & \frac{15}{2} &7 \\ \\ 12 & 10 & \frac{28}{3} \end{bmatrix}$