Explain solution for RD Sharma Class 12 Chapter Relation Exercise 1.1 Question 15 Maths textbook solution.

Answer : $R=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1),(1,3),(3,1)\}$

Hint :

A relation R on set A is

Reflexive relation: $a, b, c \in A$

If  $(a, a) \in R$ for every $a \in A$

Symmetric relation:

If $(a,b)$ is true then $(b,a)$  is also true for every $a, b \in A$

Transitive relation:

If  $(a,b)$ and , then $(b, c) \in R$  for every $(a, c) \in R$

Given :

\begin{aligned} &\text { Relation } R=\{(1,2),(2,3)\} \text { on the set }\\ &A=\{1,2,3\} \end{aligned}

Solution :

To make $R$ reflexive we will add $(1,1),(2,2),(3,3)$ to get $R^{\prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3)\}$  is reflexive.

Again, to make $R$ symmetric we will add $(3,2)$ and $(2,1)$  $R^{\prime \prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1)\}$is reflexive and symmetric .

To make $R$ transitive we will add $(1,3)$ and $(3,1)$ $R^{\prime \prime \prime}=\{(1,2),(2,3),(1,1),(2,2),(3,3),(3,2),(2,1),(1,3),(3,1)\}$ is reflexive and symmetric and transitive.