# How can I show that $ASL_n(F)$ is acting 2-transitively?

One of my friends asked me to ask this question here. This is a question from his last exam:

Let $$ASL_n(F)=\{T_{A,v}:V_n(F)\to V_n(F)\mid\exists A\in SL_n(F), \exists v\in V_n(F), T_{A,v}(x)=Ax+v\}$$ where $V_n(F)$ is a vector space of dimension $n$ over a field $F$. How can one show that $ASL_n(F)$ acts $2$-transitively on $V_n(F)$?

This is what he remembered and also something might be missing at the body of the question. Thanks for any hints!

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Consider the subgroup $T\leqslant ASL_n(F)$ generated by all the translations $t_v:x\mapsto x+v$. Obviously $T$ acts transitively on $V_n(F)$, and the stabilizer of the zero vector is $SL_n(F)$, so it suffices to prove that $SL_n(F)$ acts transitively on $V_n(F)\setminus \{0\}$.
Let $0\not= x,y\in V_n(F)$ and pick any bases $\{x_1,\ldots,x_n\}$ and $\{y_1,\ldots,y_n\}$ of $V_n(F)$ such that $x_1=x$ and $y_1=y$. Now let $M$ be the change of basis matrix defined by $Mx_i=y_i$ for all $i=1,\ldots,n$. Take $d=\text{det}(M)$ and we have that $\{y_1,\ldots,d^{-1}y_n\}$ is a basis of $V_n(F)$, so we can define another change of basis matrix $M'$ so that $M'x_i=y_i$ for $i=1,\ldots,n-1$ and $M'x_n=d^{-1}y_n$.
We have that $det(M')=1$, so $M'\in SL_n(F)$. Since $M'x=y$, and our choice of $x,y$ was arbitrary, $SL_n(F)$ acts transitively on $V_n(F)$. This completes the proof.