JEE Main 2023 — Mathematics Algebra

Given,

Set $A={1,2,3}$A=1,2,3

Now Cartesian product $A\times A={(1,1),(2,1),(1,2),........,(3,3)}$A×A=(1,1),(2,1),(1,2),........,(3,3)

Now, given the relation is reflexive,

So, $(1,1),(2,2),(3,3)\in R$(1,1),(2,2),(3,3)∈R

Also given $(1,2),(2,3)\in R,(1,3)$(1,2),(2,3)∈R,(1,3) must $\in R$∈R

Now finding, Possible cases :

Case-1: All of $(2,1),(3,2),(3,1)\notin R\rightarrow 1$(2,1),(3,2),(3,1)∈/R→1 relation.

Case-2: Only one of $(2,1),(3,2),(3,1)\in R\rightarrow 3$(2,1),(3,2),(3,1)∈R→3 relations.

For example if relation is ${(1,1),(2,2),(3,3),(2,1)}$(1,1),(2,2),(3,3),(2,1) then it is reflexive as well as transitive as $(2,2),(2,1)\rightarrow (2,1)\text{is present in relation}$(2,2),(2,1)→(2,1)is present in relation

Note that exactly two of $(2,1),(3,2),(3,1)\in R$(2,1),(3,2),(3,1)∈R is not possible because if two of these $\in R$∈R, third must $\in R$∈R to make relation transitive.

Total number of relations $=4$=4