Is $F(x) =Ax+b$ a pseudorandom function or not? Consider the following keyed function $F$: For security parameter $n,$ the
key is an $n\times n$ boolean matrix $A$ and an $n-$bit boolean vector $b$. Define
$F_{A,b} : \{0, 1\}^n-&gt;\{0, 1\}^n$ by $F_{A,b}(x) = Ax + b$, where all operations are
done modulo $2.$ Show that $F$ is not a pseudorandom function.

I have thought for a whole day but could not conquer it, so hope to get the solution here. Thanks.
I have searched the similar question in network and drawn a solution as follows:

Let's consider the distinguisher $D$ that queries its oracle $\mathcal{O}$ on arbitrary. At first, let $x=0^n$, we can get $b=\mathcal{O}(0^n)$. Then, we access the $\mathcal{O}$ with $x_1$, $x_2$, $x_1+x_2$, output $1$ if and only if $\mathcal{O}(x_1 )+\mathcal{O}(x_2 )-b=\mathcal{O}(x_1+x_2)$.


If $\mathcal{O}=F$, $Pr⁡\left[D^F{^{(\cdot)}}  (1^n)=1\right]=1$.
If $\mathcal{O}=F$ for f chosen uniformly from $Func_n$, $Pr⁡[D^{f(\cdot)}  (1^n )=1]=2^{-n}$.


The difference is $|1-2^{-n}|$, which is not negligible. Therefore, F is not a pseudorandom function.