Let A={1,2,3} and R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. Then R is 

 Reflexive: $(1,1),(2,2),(3,3)\in R\Rightarrow$ reflexive. 

 Symmetric: $(1,2)\in R$ but $(2,1)\notin R\Rightarrow$ not symmetric. 

 Transitive: check the only nontrivial chain: $(1,2)$ and $(2,3)\Rightarrow (1,3)$, which is in $R$. Pairs with $(x,x)$ keep the second pair; $(1,3)$ can only compose with $(3,3)$ giving $(1,3)$; $(2,3)$ with $(3,3)$ gives $(2,3)$. All required compositions are in $R$. $\boxed{\text{$R$ is reflexive and transitive, but not symmetric.}}$