1. Set up all required pairs

For $A=\{1,2,3\}$, a relation $R$ on $A$ is a subset of $A\times A$.

Since $R$ is reflexive, it must contain:

It is also given that $R$ contains:

So currently,

2. Use transitivity

Since $(1,2)\in R$ and $(2,3)\in R$, transitivity forces:

Thus every such relation must contain at least:

3. List the only remaining optional pairs

The full set $A\times A$ has $9$ pairs. We already know $6$ of them must be present. The remaining pairs are:

Let us decide which of these can be added while keeping transitivity, and then exclude symmetric relations.

4. Check all subsets of $\{(2,1),(3,1),(3,2)\}$

We examine all $2^3=8$ possibilities.

Case 1: Add none

Relation contains only

This is reflexive and transitive. It is not symmetric since $(1,2)\in R$ but $(2,1)\notin R$.

So this case is valid.

Case 2: Add $(2,1)$ only

Now $(2,1)$ and $(1,2)$ imply by transitivity nothing new beyond reflexive pairs, but:

• $(2,1)$ and $(1,3)$ force $(2,3)$, already present.
• $(1,2)$ and $(2,1)$ force $(1,1)$, already present.

So transitivity is preserved. It is still not symmetric because $(2,3)\in R$ but $(3,2)\notin R$.

Valid.

Case 3: Add $(3,1)$ only

Then $(2,3)$ and $(3,1)$ force:

which is not present. So transitivity fails.

Invalid.

Case 4: Add $(3,2)$ only

Then $(1,2)$ and $(2,3)$ already gave $(1,3)$. Also $(1,3)$ and $(3,2)$ force:

which is already present. And $(2,3)$ and $(3,2)$ force $(2,2)$, already present.

So transitivity holds. But check symmetry:

• $(2,3)$ and $(3,2)$ are symmetric,
• yet $(1,2)\in R$ but $(2,1)\notin R$. So relation is not symmetric.

Valid.

Case 5: Add $(2,1),(3,1)$

Since $(2,3)$ and $(3,1)$ imply $(2,1)$, already present. Since $(3,1)$ and $(1,2)$ imply:

which is missing. So transitivity fails.

Invalid.

Case 6: Add $(2,1),(3,2)$

Check transitivity:

• $(2,1)$ and $(1,3)$ imply $(2,3)$, already present.
• $(3,2)$ and $(2,1)$ imply $(3,1)$, which is missing. So transitivity fails.

Invalid.

Case 7: Add $(3,1),(3,2)$

Since $(2,3)$ and $(3,1)$ imply $(2,1)$, which is missing. So transitivity fails.

Invalid.

Case 8: Add all three: $(2,1),(3,1),(3,2)$

Then relation becomes all $9$ pairs of $A\times A$. This is reflexive, transitive, and symmetric. But we need not symmetric.

Invalid.

5. Count valid relations

Valid cases are:

1. none added
2. only $(2,1)$
3. only $(3,2)$

Hence the number of required relations is:

6. Compare with stored answer

Derived answer = $3$. Stored correct answer = $3$. They agree.