# Commutative pairs of relations do not define an equivalence relation

I don't understand the following question, please take a look and help if you can:

Let $M$ be the set of relations over $A=\{1,2,3\}$

a) State the number of elements in $M$. Since $|A|=3$, then the number of relations in $M$ is $2^9$?

b) (The main question I don't understand) We define a relation $S$ over $M$ so that for every $R_1,R_2 \in M$: $(R_1,R_2) \in S$ iff $R_1R_2=R_2R_1$. Prove or disprove if $S$ is an equivalence relation over $M$.

(Regarding b and what i tried to do)

Equivalence relation is defined only if the relation holds reflexivity, symmetry, and transitivity properties. And $R_1R_2=R_2R_1$ is the symmetry property, but it doesn't hold here though I don't know how to write it mathematically. I know it's false, but I'm not sure how to disprove it correctly, showing $S$ doesn't hold symmetry properties.

You are correct that $S$ is not an equivalence relation. However, $S$ is symmetric. Indeed, if $(R_1,R_2)\in S$, then $R_1R_2=R_2R_1$ implies $R_2R_1=R_1R_2$ and hence $(R_2,R_1)\in S$.
The issue is that $S$ is not transitive. If it were, then $R_1R_2=R_2R_1$ and $R_2R_3=R_3R_2$ would imply $R_1R_3=R_3R_1$. Let's find a counterexample to this.
Set $R_1:=\{(1,1)\}$, $R_2:=\{(2,2)\}$, and $R_3:=\{(1,3)\}$. Then $$R_1R_2=R_2R_1=R_2R_3=R_3R_2=\emptyset$$ so that $(R_1,R_2),(R_2,R_3)\in S$. However $R_1R_3=\{(1,3)\}\ne\emptyset=R_3R_1$. Therefore $(R_1,R_3)\not\in S$.