Show that the number of equivalence relations on the set $A = \{1, 2, 3\}$ containing $(1, 2)$ and $(2, 1)$ is $2$.

(B) Let $A = \{1, 2, 3\}$. An equivalence relation $R$ must be reflexive,symmetric,and transitive.
Since $R$ contains $(1, 2)$ and $(2, 1)$,by symmetry it must contain $(1, 1)$ and $(2, 2)$ to be reflexive on these elements,and $(3, 3)$ must be included for reflexivity on $3$.
Thus,the smallest equivalence relation $R_1$ containing $(1, 2)$ and $(2, 1)$ is $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.
To form another equivalence relation,we must add elements while maintaining transitivity. If we add $(2, 3)$,then by symmetry we must add $(3, 2)$. For transitivity,since $(1, 2) \in R$ and $(2, 3) \in R$,we must have $(1, 3) \in R$. By symmetry,we must also have $(3, 1) \in R$.
Adding these pairs results in the universal relation $R_2 = A \times A = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)\}$.
Thus,there are exactly $2$ such equivalence relations.