How many "transitive relation" can be formed by A×A If A is non-empty set then how may " transitive relation " can be made by A×A ?
 A: There is no simple way to get a solution but you can interpretate your problem in this way:
For each relation $\sim$ of $A$ we can define the map 
$\psi_\sim: A\to \mathcal{P}(A)$ 
such that 
$\psi_\sim(a):=\{b\in A: a\sim b\}$
In this case you have that if $\sim$ is transitive then  for each $b\in \psi_\sim(a)$ 
$\psi_\sim(b)\subseteq \psi_{\sim}(a)$
So 
$\{\sim : \sim transitive \}\cong \Lambda$
where $\Lambda:= \{\psi:A\to \mathcal{P}(A): \forall a,b \  if  \ b\in \psi(a)  \ then  \ \psi(b)\subseteq \psi(a)\}$
Now we want prove to determine the cardinality of $\Lambda$. For simplicity 
$A=\{1,\dots , n\}$
We suppose that $\psi(i)=A$ for each $i< n$ then the only choice of $\psi(n)$  to get  $\psi$ transitive can be $\psi(n)=\{n\}$ or $\psi(n)=A$. So in this case we have 
$|\{\psi\in \Lambda : \psi(i)=A \forall i< n\}|=2$
We suppose that $\psi(n-1)\neq A$ while $\psi(i)=A$ for each $i<n-1$ . Then $\psi(n-1)$ does not contain $1,\dots ,n-2$ because otherwise $A=\psi(i)\subseteq \psi(n-1)$ and it is not possible. So we can have only 
$\psi(n-1)=\{n-1\}$ or $\psi(n-1)=\{n-1, n\}$ or $\psi(n-1)=\{n\}$
In the first case we can choice $\psi(n)=\{n\}$  or $\psi(n)=\{n, n-1\}$ or $\psi(n)=A$ while in the second case  we can have 
$\psi(n)=\{n\}$ or $ \psi(n)=\{n,n-1\}$
and in the third case we have 
$\psi(n)=\{n\}$
So we have 6 cases.
