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

$A= \begin{bmatrix} 0 &-\frac{1}{3} &-\frac{1}{2} \\ \\ \frac{1}{3} &0 & -\frac{1}{5}\\ \\ \frac{1}{2} & \frac{1}{5} &0 \\ \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix}$

Given: Here  $a_{ij}= \frac{i-j}{i+j}$

Here we have to construct $4\times 3$  matrix according to $\frac{i-j}{i+j}$

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

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

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

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

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

$\! \! \! \! \! \! \! \! \! a_{11}=\frac{1-1}{1+1}= \frac{0}{2}= 0 \\\\a_{12}=\frac{1-2}{1+2}=- \frac{1}{3} \\\\a_{13}= \frac{1-3}{1+3}= -\frac{2}{4}= \frac{1}{2}$                           $\! \! \! \! \! \! \! \! \! a_{21}=\frac{2-1}{2+1}= \frac{1}{3} \\\\a_{22}=\frac{2-2}{2+2}= \frac{0}{4}= 0 \\\\a_{23}= \frac{2-3}{2+3}= -\frac{1}{5}$

$\! \! \! \! \! \! \! \! \! a_{31}=\frac{3-1}{3+1}= \frac{2}{4} = \frac{1}{2}\\\\a_{32}=\frac{3-2}{3+2}= \frac{1}{5} \\\\a_{33}= \frac{3-3}{3+3}= \frac{0}{6}= 0$                             $\! \! \! \! \! \! \! \! \! a_{41}=\frac{4-1}{4+1}= \frac{3}{5} \\\\a_{42}=\frac{4-2}{4+2}= \frac{2}{6}= \frac{1}{3} \\\\a_{43}= \frac{4-3}{4+3}= \frac{1}{7}$

Substituting these values in Matrix $A$ , we get

$A= \begin{bmatrix} 0 &-\frac{1}{3} &-\frac{1}{2} \\ \\ \frac{1}{3} &0 & -\frac{1}{5}\\ \\ \frac{1}{2} & \frac{1}{5} &0 \\ \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix}$