A guess about invertible sign matrix

Question

For any $n\times n$ invertible matrix $A＝(a_{ij})$, which satisfies $a_{ij}\in\{0,1,-1\}$. Assume $A^{-1}＝(b_{ij})$.

I guess $\vert b_{ij}\vert \le 1$, because I have found it is right when $n＝1,2$.

Is this right or wrong? If wrong, can you give me a counterexample？

Thanks for the counterexample.

But i wonder can we prove that $\max_{i}\sum_{j＝1}^{n}\vert b_{ij}\vert\le n$？

Thanks again.

Answer 1

Note that $$\begin{bmatrix} 1&1&1&-1\\ 1&1&1&1 \\1&0&1&1\\0&1&1&1\end{bmatrix}^{-1}= \begin{bmatrix} 0&1&0&-1\\ 0&1&-1&0 \\ \frac12&-\frac32&1&1\\ -\frac12&\frac12&0&0 \end{bmatrix}.$$

EDIT: For the second question,

$$\begin{bmatrix} 1&1&1&1&-1\\ 1&1&1&1&1 \\ 1&1&0&1&1\\ 1&0&1&1&1 \\ 0&1&1&1&1 \end{bmatrix}^{-1} = \begin{bmatrix} 0&1&0&0&-1\\ 0&1&0&- 1&0\\ 0&1&-1&0&0\\ \frac12&-\frac52&1& 1&1\\ -\frac12&\frac12&0&0&0\end{bmatrix}. $$