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> Suppose $A(p, n)=(a\_{ij}(p))\_{i, j \leq n}$ is an $n\times n$ random matrix over $\mathbb{F\_2}$, with all its entries being i.i.d. and such that $P(a\_{ij}(p) = 1) = p$, where $p$ is some real number from $[0; 1]$. What is the largest possible probability, that $A(p, n)$ is non-singular and with what $p$ is it reached?
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Note, that $A(p, n)$ is non-singular iff $\det(A(p, n)) = 1$.


Solution for $n=1$:


$\det(A(p, 1)) = 1$ with probability $p$. The maximum of $\det(A(p, 1))$ is $1$ and it is reached with $p = 1$.


Solution for $n = 2$:


$\det(A(p, 2)) = 1$ with probability $2p^2(1 - p^2)$. The maximum of $P(\det(A(p, 2))=1)$ is $\frac{1}{2}$ and it is reached with $p = \frac{1}{\sqrt{2}}$.


However, I would like to know some sort of general formula (or at least asymptotics).


After I failed to solve this problem using determinants, I tried to prove this using the fact that a square matrix is non-singular iff its rows are linearly dependent. As there exists only one non-zero element in $\mathbb{F\_2}$, we can write linear dependence of the vector system $\{v\_i\}\_{i \leq n}$ in $\mathbb{F}\_2^n$ as $\forall S \subset \{1, ... , n\}$ such that $S \neq \emptyset$ we have $\sum\_{i \in S} v\_i \neq \overline{0}$. I know the probability that a given set of vectors with i.i.d. random entries Bernoulli distributed with parameter $p$ $\{v\_i\}\_{i \leq k}$ over $\mathbb{F}\_2^n$ satisfy $\sum\_{i = 1}^k v\_i \neq \overline{0}$ is $(1 - \frac{p((1 - 2p)^k - 1)}{1 - 2p})$. However, I do not know how to proceed further in this direction.


[This question on MSE](https://math.stackexchange.com/q/3456885/407165)



What is the largest possible probability that a random matrix over $\mathbb{F}_2$ is non-singular?
For $n\rightarrow\infty$ the probability ${\cal P}\_\infty$ that $A(p,n)$ is nonsingular becomes independent of $p\in(0,1)$, given by 
$${\cal P}\_\infty=\prod\_{i=1}^\infty(1-2^{-i})=0.2887880951$$
See theorem 3.2 in [Properties of random matrices and applications](http://www.cs.toronto.edu/~cvs/coding/random_report.pdf) (2007).


For finite $n$, there is a result that could be instructive, which is the expectation value $E(p,n)$ of the number of linear dependencies among the rows of $A(p,n)$. This is given by theorem 4.5,
$$E(p,n)=2^{-n}\sum \_{j=1}^n \binom{n}{j} \left(1+(1-2 p)^j\right)^n.$$
A plot of $E(p,n)$ as a function of $p$ becomes flatter and flatter with increasing $n$ (see below for $n$ up to 20), consistent with the understanding that for large $n$ the probability that the matrix is singular no longer depends on $p$.


![](https://ilorentz.org/beenakker/MO/f2singular.png)


