About a curious cross-product/determinant identity Whilst proving the fact that one definition of area for a domain inside a parameterisation of some surface embedded in $\mathbb{R}^3$ is well defined, my lecturer made a claim "by linear algebra" that amounts to:

If $A$ is a $3$ by $2$ matrix and $B$ is a $2$ by $2$ matrix, $ABv\wedge ABw = \det{B}(Av\wedge Aw)$ for any vectors $v,w \in \mathbb{R}^2$.

(The claim was actually made for $v=e_1,w=e_2$, but this result is equivalent to the one above (I think).)
I spent a long time trying to prove it using things like the Levi-Civita symbol, and was unsuccessful. I eventually did prove it by resorting to writing $B$ as a $2$ by $2$ matrix and doing the explicit calculation, but this felt unsatisfactory, so my question is:

Is there less computation-based method of proving this? Even better, is there a more general result that this is a special case of?

I'm curious because it looks like a simple vector calculus formula that should be easy to prove without calculation, and the links between determinants, cross products and volume make me think that there should be some kind of more general version. On the other hand it may be true because of the link to differential geometry and so doesn't generalise...
 A: Recall that, for any $n\times n$ matrix $B$ and any vectors $v_1,\ldots,v_n \in \mathbb{R}^n$, we have
$$
(Bv_1) \land \cdots \land (Bv_n) \;=\; (\det B)(v_1\land \cdots \land v_n)
$$
(See here on Wikipedia.)
Moreover, it is easy to check that, for any $m\times n$ matrix $A$ and any $k\leq \min(m,n)$, the function $\bigwedge^k \!A\colon \bigwedge^k \mathbb{R}^n \to \bigwedge^k \mathbb{R}^m$ defined by
$$
(\textstyle\bigwedge^k \!A)(v_1\land \cdots \land v_k) \;=\; (Av_1) \land \cdots \land (Av_k)
$$
is well-defined and linear.
If $A$ is a $3\times 2$ matrix, $B$ is a $2\times 2$ matrix, and $v,w\in\mathbb{R}^2$, it follows that
$$
\begin{align*}
(ABv) \land (ABw) \;&=\; (\textstyle\bigwedge^2 \!A)(Bv\land Bw) \\[3pt]
&=\; (\textstyle\bigwedge^2 \!A)\bigl((\det B)(v\land w)\bigr) \\[3pt]
&=\; (\det B)(\textstyle\bigwedge^2 \!A)(v\land w) \\[3pt]
&=\; (\det B)(Av\land Aw).
\end{align*}
$$
The same argument shows that
$$
(ABv_1) \land \cdots \land (AB v_n) \;=\; (\det B)(Av_1 \land\cdots\land Av_n)
$$
for any $v_1,\ldots,v_n\in\mathbb{R}^n$, any $m\times n$ matrix $A$ (where $m\geq n$), and any $n\times n$ matrix $B$.
