I do not think that boolean product of two matrices take $O(n^2)$. Rather boolean product of two matrices takes $O(n^3)$ just like usual matrix multiplication, Now if $M$ is the incidence matrix of the relation, then $M^k$ an entry $1$ in those $[i,j]$ cells where there is a path of length $k$. So find transitive closure we find $M^n$ where $n$ is the number of elements in the set $A$ on which the relation is defined. So we need a total of $O(n^4)$ operations in a usual method. But Warshall’s Algorithm uses a dynamic programming approach to reduce this running time to $O(n^3)$.