<s>[INST] Finding the probability that the matrix has full rank 
There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus,
  only $16$ possible $2\times 2$ matrices over $Z_2$, and $6$ of them
  have full rank:
$\begin{pmatrix}0&1\\   1&0\end{pmatrix}  \quad \begin{pmatrix}1&1\\  
> 1&0\end{pmatrix}  \quad \begin{pmatrix}0&1\\   1&0\end{pmatrix}  \quad
> \begin{pmatrix}0&1\\   1&1\end{pmatrix}  \quad \begin{pmatrix}1&1\\  
> 0&1\end{pmatrix}  \quad \begin{pmatrix}1&0\\   0&1\end{pmatrix}$
Randomly generate a $n\times n$ matrix over $\mathbb {Z}_2$ (where n
  is big, say, $1000$). What's the probability that the matrix has full
  rank?

I'd want to talk something regarding to this question. 
I found the probability is equal to:
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^n} \tag{1}$
On other hand, what's difference between $2^n$ and $2^{n^2}$?
$\dfrac{(2^n-1)(2^n-2)(2^n-2^2)...(2^n-2^{n-1})}{2^{n^2}} \tag{2}$
Can you tell whether or not I'm wrong? 
Regards!
 [/INST] The second formula seems correct to me.  The rows of the matrix must be linearly independent.  The first row can be any vector but the zero vector, so there are $2^n-1$ choices.  Suppose that $k$ linearly independent rows have been added.  The next row can be any vector not in the linear span of the the first $k$ rows.  Since these form a $k-$dimensional vector space over GF(2), there are $2^n-2^k$ choices for row $k+1.$
Since there are $2^{n^2}$ possible $n\times n$ matrices, the denominator should be $2^{n^2}$.
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