The best and the most reliable order to satisfy properties of equivalence relation is in the given order
=> Reflexive Closure-->Symmetric Closure-->Transitivity closure
The reason for this assertion is that like for instance if you are following the order
=> Transitivity closure-->Reflexive Closure-->Symmetric Closure
(just like has been asked) ,you may just end up with elements that you added for symmetric closure not being accounted for transitivity as has been shown in the example given in question which has been cited here for reference
R={(2,1),(2,3)} .
Transitive closure: {(2,1),(2,3)}.
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}.
Since the set is missing (1,3) and (3,1) to be transitive, it is not an equivalence relation.
But if you follow the order of satisfying Reflexive Closure first,then Symmetric Closure and at last Transitivity closure,then the equivalence property is satisfied as shown.
R={(2,1),(2,3)} .
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}
Transitive closure:{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,2),(3,1),(3,3)}
To be transitive, $xRz$ needs to hold whenever you have $x$, $y$, and $z$ such that $xRy$ and $yRz$. You're only testing the case $x=y=1$ and $z=3$, but there might be other cases. For instance, in your first example, you can let $x=1$, $y=3$, and $z=2$, and you find that $1R3$ and $3R2$, but $1R2$ is not true. So the relation is not transitive.
(By the way, you should not call that first relation an "equivalence relation", since an equivalence relation is required to be transitive, in addition to being reflexive and symmetric.)
Best Answer
You did not check all cases. It is not transitive because both $(1,3)$ and $(3,2)$ belong to it, whereas $(1,2)$ doesn't.