I am probably a bit lost but I found the following 'cross-product' property in a paper :
$(Ma) \times (Mb) = \frac{1}{det([a,b]^{-1})} ( M_1 \times M_2 ) $
With $M$ a $3\times2$ Matrix, $a$ and $b$ two $2\times1$ vector. $[a,b]$ is a $2\times2$ matrix obtained by concatenating (horizontaly)the two vector a and b and $M_1$, $M_2$ correspond to the 1st and 2nd column of $M$.
I am currently struggling a lot to prove this formula and would like to know if someone can help me. I m sorry for this question which is probably quite trivial but I really don t see where to start.
Thank you very much for your help.
F
Best Answer
Let the entries of $a$ be $a_1$ and $a_2$ and those of $b$ be $b_1$ and $b_2$. Then the left hand side reads
$$(a_1M_1 + a_2M_2) \times (b_1M_1 + b_2M_2).$$
Using the distributive property we know from ordinary multiplication, this expands to
$$a_1b_1 (M_1 \times M_1) + a_1b_2 (M_1 \times M_2) + a_2b_2 (M_2 \times M_2) + a_2b_1 (M_2 \times M_1)$$
Ignoring terms of the form $X \times X$ (because they are zero) and using the fact that $M_2 \times M_1 = - M_1 \times M_2$ we end up with:
$$(a_1b_2 - a_2b_1)(M_1 \times M_2)$$
This is already almost what we wanted: obviously the scalar $(a_1b_2 - a_2b_1)$ appearing here equals $\det[a, b]$ so all we need is the general fact that $\det A = \frac{1}{\det A^{-1}}$ that holds for all invertible matrices $A$.
This answers the question.
Notice that I tried to avoid actually computing any cross products and worked with their properties (bilinearity, anti-symmetry) instead. This is often a good idea when cross products are involved.