Let $A \in \mathbb{R}^{b \times b}$ be a matrix with non-negative entries. Suppose that $A$ is invertible. Further suppose that all the diagonal entries of $A$ are non-null. My claim is the following: $$\text{bin}(A^{-1})=\text{bin}(A^{b-1})\,,$$ where $\text{bin}: \mathbb{R}^{b \times b}\rightarrow \{0,1\}^{b \times b}$ gives the binary sparsity pattern of $A$, in the sense that an entry of matrix $\text{bin}(A)$ is $1$ if and only if the corresponding entry of $A$ is different from $0$. For instance $$\text{bin}\left(\begin{bmatrix}1&2&0\\0&2&1\\1&0&2\end{bmatrix}^{-1}\right)=\text{bin}\left(\begin{bmatrix}1&2&0\\0&2&1\\1&0&2\end{bmatrix}^2\right)=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}\,.$$
Notice that, by Cayley-Hamilton's theorem and the fact that the diagonal of $A$ is full, I expect any counter-example to my claim to be such that one entry of $\text{bin}(A^{-1})$ is $0$ while the same entry of $\text{bin}(A^{b-1})$ is $1$.
$$ \pmatrix{1 & 1 & 1\cr 1 & 2 & 2\cr 1 & 2 & 3}^{-1} = \pmatrix{2 & -1 & 0\cr -1 & 2 & -1\cr 0 & -1 & 1\cr}$$