Number of Relations that satisfy a condition

Question

This is a multiple choice question from my Text Book

Let $A=\{1,2,3\}$. The no. of relations containing $(1,2)$ and $(1,3)$ which are reflexive and Symmetric but not transitive is

(A) $1$

(B) $2$

(C) $3$

(D) $4$

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My Approach: $A=\{1,2,3\}$

Relation $R$ must contain $(1,2)$ and $(1,3)$

For $R$ to be Reflexive, it must contain $(2,2)$ and $(1,1)$

For $R$ to be symmetric, it must contain $(2,1)$ and $(3,1)$

For $R$ to not to be Transitive, it must not contain $(2,3)$ and $(3,2)$

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

Anyother addition to $R$ will not satisfy the stated condition.

Hence, option $A$ is correct

Am I right?

[Edit:

R contains $(3,3)$ as well]

Answer 1

You must also have $(3,3)$ for the relation to be reflexive. Because you have $(2,1)$ and $(1,3)$ you would need $(2,3)$ for the relation to be transitive. Similarly, you need $(3,2)$. If either or both of these are missing, the relation is not transitive, so there would be $3$ choices, but the requirement of symmetry says they have to both be missing and there is only one choice.

Answer 2

For R to be Reflexive, it must contain (2,2) and (1,1)

It also must contain $(3,3)$

For R to not to be Transitive, it must not contain (2,3) and (3,2)

That is not true.