We are given a relation $R$ on $\mathbb{Z} \times \mathbb{Z}$ defined by $(a,b)R(c,d) \iff 5 \mid (ad-bc).$ We must check whether $R$ is reflexive, symmetric, and transitive.

1. Reflexive

A relation is reflexive if every element is related to itself.

Take any $(a,b) \in \mathbb{Z} \times \mathbb{Z}$. Then $(a,b)R(a,b) \iff 5 \mid (ab-ba)=0.$ Since $0$ is divisible by $5$, this is always true.

So, $R$ is reflexive.

2. Symmetric

A relation is symmetric if $(a,b)R(c,d) \implies (c,d)R(a,b).$

Assume $(a,b)R(c,d)$. Then $5 \mid (ad-bc).$ Now, $cb-da = -(ad-bc).$ If $ad-bc$ is divisible by $5$, then its negative is also divisible by $5$. Hence $5 \mid (cb-da),$ which means $(c,d)R(a,b).$

So, $R$ is symmetric.

3. Transitive

A relation is transitive if $(a,b)R(c,d) \text{ and } (c,d)R(e,f) \implies (a,b)R(e,f).$

We test this by finding a counterexample.

Take $(a,b)=(1,0), \quad (c,d)=(0,0), \quad (e,f)=(0,1).$

Check $(1,0)R(0,0)$

We compute $1\cdot 0 - 0\cdot 0 = 0,$ and $0$ is divisible by $5$. So $(1,0)R(0,0).$

Check $(0,0)R(0,1)$

We compute $0\cdot 1 - 0\cdot 0 = 0,$ and $0$ is divisible by $5$. So $(0,0)R(0,1).$

Check whether $(1,0)R(0,1)$

We compute $1\cdot 1 - 0\cdot 0 = 1.$ But $1$ is not divisible by $5$. Hence $(1,0) \not R (0,1).$

Thus transitivity fails.

So, $R$ is not transitive.

4. Final conclusion

The relation $R$ is:

• Reflexive
• Symmetric
• Not transitive

Therefore, the correct option is $\boxed{\text{B: Reflexive and symmetric but not transitive}}.$

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.