Prove $(Ma)\times(Mb)=(\det M)(M^{-1})^T(a\times b)$ for matrix transformation $M$.

Question

I wanted to prove $(Ma)\times(Mb)=(\det M)(M^{-1})^T(a\times b)$ where the formula could be found here wiki's cross product page However, I had a hard time counting the index, and I was wondering if there's any easier way to prove it.

Answer 1

They are equal because \begin{aligned} \left[(\det M)(M^{-1})^T(a\times b)\right]\cdot Mc &=(\det M)(a\times b)\cdot M^{-1}Mc\\ &=\det(M)(a\times b)\cdot c\\ &=\det(M)\det[a,b,c]\\ &=\det[Ma,Mb,Mc]\\ &=(Ma\times Mb)\cdot Mc \end{aligned} for every vector $c$.

Answer 2

By the definition of the determinant, $\epsilon_{ijk}M_{il}M_{jm}M_{kn}=(\det M)\epsilon_{lmn}$. Multiplying both sides by $a_lb_m$, $[M^T(Ma\times Mb)]_n=(\det M)(a\times b)_n$, i.e. $M^T(Ma\times Mb)=(\det M)a\times b$. The desired result then follows from $(M^T)^{-1}=(M^{-1})^T$.