It is known that the entries of the inverse of an (invertible) $n \times n$ zero-one matrix $A$ can grow exponentially in $n$. See this MO question:
[Bounding the absolute sum of entries of the inverse of a 0-1 matrix](https://mathoverflow.net/questions/81496/bounding-the-absolute-sum-of-entries-of-the-inverse-of-a-0-1-matrix?lq=1).


What if $A$ is known to have constant column sums (so a multiple of a stochastic matrix)? Or, more generally, what if we have bounds on the entries of $A^\top A$?


In the first question (with no additional constraints on $A$), Noam Elkies gives an interesting example in which $A\_{ij}=1$ if and only if $j=i$, $j=i+1$ or $j=i+3$. Interpreting these (mod $n$) leads to only three more ones in the (now circulant) matrix, yet the inverse becomes well-behaved.



Bounding entries of the inverse of certain zero-one matrices
Let $A$ be a matrix with column sums at most $3$ with large elements in the inverse matrix. Construct a matrix of the form $\begin{pmatrix}A&0\\B&C\end{pmatrix}$, where $B$ is a $4\times n$ matrix and $C$ is a $4\times 4$ matrix. You may do this so that the resulting matrix has constat column sums (equal to $3$) and is non-degenerate. The inverse of this matrix contains $A^{-1}$ in the upper-left corner.


