Bounding entries of the inverse of a matrix with bounded entries Let $A$ be an $n$-by-$n$ matrix with integer entries whose absolute values are bounded by a constant $C$. It is well-known that the entries of the inverse $A^{-1}$ can grow exponentially on $n$. (See the replies to Estimations for the size of the biggest entry in an inverse Matrix .) Can they grow more rapidly than exponentially? Or are they bounded by $(C+O(1))^n$?
 A: They can grow more rapidly. We assume that $C=1$.
Take $n=2^{ k+2}$ where $k$ is a positive integer and let 
EDIT. there is a mistake of calculation below.
$A=\begin{pmatrix}1&0&0&0\\0&I&0&0\\0&0&I&0\\H&0&0&1\end{pmatrix}$ be a  matrix of dimension $p=3n/4+1$, $H$ be a Walsh Sylvester matrix of dimension $n/4$ and $I$ be the identity matrix of dimension $n/4$.
Then $A^{-1}=Adjoint(A)$.
Note that $|{A^{-1}}_{p,1}|= \det(H)= (n/4)^{n/8}$ (cf. case of equality in the Hadamard's inequality).
Conversely, let $A\in M_n(\mathbb{Z})$ with $C=1$.  $|{A^{-1}}_{i,j}|\leq {Adjoint(A)}_{i,j}$ which is a det of a minor of $ A$. According to the Hadamard's inequality, such a minor is $\leq {(n-1)}^{(n-1)/2}$. We deduce the following
$\textbf{Proposition}$. Under the above conditions, $\log(|{A^{-1}}_{i,j}|)=O(n\log(n))$ and we cannot do better.
A: The inverse of a nonsingular square $n \times n$ matrix containing only 0s and 1s can be as large as $n^{n/2(1 + o(1))}$, and can be no larger. This is proved in:
N. Alon and V. H. Vu, Anti-Hadamard matrices, coin weighing, threshold gates and indecomposable hypergraphs, J. Combinatorial Theory, Ser. A 79 (1997), 133-160.
Credit here: https://mathoverflow.net/a/182393/92003
