Original problem:
We have invertible matrix $M$ dimension $n$ over the $\mathbb F_2$. If $M$ contains column of all ones, then there exists two different rows $V,W$ of $M$ such that: $$<V,W>=\sum_{i}v_iw_i=1. $$
After sometime and with the help of my friend we find out that the problem is equal to the $$M*M^T\neq I,$$ where $I$ is the identical matrix.
And this problem can be solved by solving this one:
We have invertible matrix $M$ dimension $n$ over the $\mathbb F_2$. If $M$ contains column of all ones, then it is not orthogonal matrix.
With $\mathbb F_2$ we denote the field with two elements: zero and one.
Every suggestions and ideas are welcome.
Edit: It's all ok when $n$ is even, but the odd $n$'s are problem.
Suppose $V$ is the row of all $1$'s, and $W$ is any other row. Then $V\cdot W = 0$ implies that $W$ must contain an even number of entries which are non-zero. But then $W \cdot W = 0$.