Let $$R$$ be a relation on $$Z \times Z$$ defined by $$(a, b)R(c, d)$$ if and only if $$ad - bc$$ is divisible by $$5$$. Then $$R$$ is

The relation $$R$$ on $$\mathbb{Z} \times \mathbb{Z}$$ is defined by $$(a, b)R(c, d)$$ if and only if $$ad - bc$$ is divisible by 5.

Check Reflexivity.

$$(a, b)R(a, b) \iff ab - ba = 0$$ is divisible by 5. $$\checkmark$$

$$R$$ is reflexive.

Check Symmetry.

If $$(a, b)R(c, d)$$, then $$5 \mid (ad - bc)$$.

We need to check: is $$5 \mid (cb - da)$$?

$$cb - da = -(ad - bc)$$. Since $$5 \mid (ad - bc)$$, we have $$5 \mid (-(ad-bc)) = cb - da$$. $$\checkmark$$

$$R$$ is symmetric.

Check Transitivity.

We need: if $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$, then $$(a,b)R(e,f)$$.

Given: $$5 \mid (ad - bc)$$ and $$5 \mid (cf - de)$$.

Need to show: $$5 \mid (af - be)$$.

Counterexample:

Take $$(a, b) = (1, 1)$$, $$(c, d) = (5, 5)$$, $$(e, f) = (1, 2)$$.

Check $$(1,1)R(5,5)$$: $$ad - bc = 1 \cdot 5 - 1 \cdot 5 = 0$$. Divisible by 5. $$\checkmark$$

Check $$(5,5)R(1,2)$$: $$cf - de = 5 \cdot 2 - 5 \cdot 1 = 5$$. Divisible by 5. $$\checkmark$$

Check $$(1,1)R(1,2)$$: $$af - be = 1 \cdot 2 - 1 \cdot 1 = 1$$. NOT divisible by 5. $$\times$$

So transitivity fails.

$$R$$ is reflexive and symmetric but not transitive.

The correct answer is Option (1): $$\boxed{\text{Reflexive and symmetric but not transitive}}$$.