I'm trying to understand why the following is incorrect.
Given a $STCON$ problem, specifically a graph and nodes $(G, s, t)$, we can assume we are given it's adjacency matrix, $A$. By adding self-loop (filling the main diagonal with $1$'s) we can multiple this matrix with itself n times, each time checking if the $s,t$ entry is $1$, and if so returning true.
We know multiplying binary matrices is in $L$ (need to "save" only couple of indices), and by also saving $n$ (number of total matrix multiplications) one can solve $STCON$ also in $L$.
Where am I wrong here?