Prove that a matrix is invertible?

Question

Let $A_{20 \times 20}$ be a real matrix such that:
$\ \ \ \bullet$ $a_{ii}=0$ for $1 \le i \le 20$
$\ \ \ \bullet$ $a_{ij} \in \{-1;1\}$ for $1 \le i,j \le 20$ and $ i \neq j$
Prove that $A$ is nonsingular.

Anyone can help me find out an useful idea for this problem.

Answer 1

HINT:

$$A^2 = I_{20}\ (\!\! \! \! \!\mod 2\,)$$

Answer 2

I think you are aware that

$$\det A=\sum_{\sigma}\prod_{i=1}^{20} a_{i\sigma(i)}$$

The number of $\sigma$ such that $\forall i:\sigma(i)\ne i$ is $!20$ (http://en.wikipedia.org/wiki/Derangement) which is an odd number (it can be proven by induction that $!n=n-1(\operatorname{mod}2)$). therefore

$$\det A=1(\operatorname{mod}2)$$ so $\det A\ne0$. Direct proof I think.