Matrix Identity Proof Let $A$ and $C$ be $3 \times 2$ matrices and let $B$ be a $2 \times 2$ matrix such that $AB=C$. Prove that:
$$||A_1 \times A_2 || \cdot |\det B| = ||C_1 \times C_2 ||$$
where $A_i$ and $C_i$ are the $i$th columns of $A$ and $C$.
 A: Write $A_n=(a_{1n},a_{2n},a_{3n})$, $C_k=(c_{1n},c_{2n},c_{3n})$ and recall
$$
A_1\times A_2= \left|\matrix{i&j&k\\ a_{11}&a_{21}&a_{31}\\a_{12}&a_{22}&a_{32}} \right|= \left|\matrix{i&a_{11}&a_{12}\\ j&a_{21}&a_{22}\\k&a_{31}&a_{32}} \right|.
$$
Now observe that
$$
\left( \matrix{i&a_{11}&a_{12}\\ j&a_{21}&a_{22}\\k&a_{31}&a_{32}}\right)\cdot \left(\matrix{1&0\\0&B }\right)=\left( \matrix{i&c_{11}&c_{12}\\ j&c_{21}&c_{22}\\k&c_{31}&c_{32}}\right)$$
follows from the assumption $AB=C$.
Now take the determinant and use its multiplicity to get
$$\det\left( \matrix{i&a_{11}&a_{12}\\ j&a_{21}&a_{22}\\k&a_{31}&a_{32}}\right)\det \left(\matrix{1&0\\0&B }\right)=\det\left( \matrix{i&c_{11}&c_{12}\\ j&c_{21}&c_{22}\\k&c_{31}&c_{32}}\right).
$$
Thus we have

$$
(A_1\times A_2)\det B=C_1\times C_2.
$$

Finally, take the norm to obtain the desired formula.
A: This is the same answer as Julien's above, with the $i,j,k$ replaced by a linear functional equivalent.
The cross product $x \times y$ can be defined as the unique vector that satisfies $\langle z, x \times y \rangle = \det \begin{bmatrix} x & y & z\end{bmatrix}$.
If $z$ is an arbitrary vector, then $\begin{bmatrix} a_1 & a_2  & z\end{bmatrix} \begin{bmatrix} B & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} A  & z\end{bmatrix} \begin{bmatrix} B & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix} Ab_1 & A b_2  & z\end{bmatrix} = \begin{bmatrix} c_1 & c_2  & z\end{bmatrix}$.
This gives $\langle z, c_1 \times c_2 \rangle = \det \begin{bmatrix} c_1 & c_2 & z\end{bmatrix} = \det \begin{bmatrix} a_1 & a_2  & z\end{bmatrix} \det \begin{bmatrix} B & 0 \\ 0 & 1\end{bmatrix} =  \langle z, (\det B)(a_1 \times a_2) \rangle$.
Uniqueness gives the desired result.
