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 $\mathrm{R}$ is

Given relation $R : \left(a,b\right) R \left(c,d\right)$

And $ad-bc=5k \left\{\text{where }k\in Z\right\}$

Now, for reflexive $R : \left(a,b\right) R \left(a,b\right)$

$\Rightarrow ab-ba=0$ which is divisible by $5$

So, the relation is reflexive,

Now, for symmetric $R : \left(c,d\right) R \left(a,b\right)$

$\Rightarrow$ $bc-ad=-5k$ which is divisible by $5$

So, the relation is symmetric,

Now for transitive relation, $R : \left(a,b\right) R \left(c,d\right) R \left(e,f\right)$

Taking example $\left(a,b\right)\equiv \left(1,2\right), \left(c,d\right)\equiv \left(10,10\right), \left(e,f\right)\equiv \left(4,4\right)$

Now, $ad-bc=10-20=-10$

$cf-ed=40-40=0$

But $af-be=8-4=4$ which is not divisible by $5$,

Hence, the relation is reflexive, symmetric but not transitive.