# How can one prove the following identity of the cross product? ( M a ) &#x00D7;<!--

How can one prove the following identity of the cross product?
$\left(Ma\right)×\left(Mb\right)=det\left(M\right)\left({M}^{\mathrm{T}}{\right)}^{-1}\left(a×b\right)$
$a$ and $b$ are 3-vectors, and $M$ is an invertible real $3×3$ matrix.
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Tatiana Gentry
Recall that $u×v$ is the only vector $w\in {\mathbb{R}}^{3}$ that satisfies$\mathrm{\forall }\phantom{\rule{thinmathspace}{0ex}}x\in {\mathbb{R}}^{3},\phantom{\rule{1em}{0ex}}det\left(u,v,x\right)=⟨w,x⟩$So, we have$\begin{array}{rl}det\left(M\right)⟨a×b,x⟩& =det\left(M\right)det\left(a,b,x\right)\\ & =det\left(Ma,Mb,Mx\right)\\ & =⟨Ma×Mb,Mx⟩\\ & =⟨{M}^{T}\left(Ma×Mb\right),x⟩\end{array}$Thus$\mathrm{\forall }\phantom{\rule{thinmathspace}{0ex}}x\in {\mathbb{R}}^{3},\phantom{\rule{1em}{0ex}}⟨det\left(M\right)\left(a×b\right),x⟩=⟨{M}^{T}\left(Ma×Mb\right),x⟩$It follows that $det\left(M\right)\left(a×b\right)={M}^{T}\left(Ma×Mb\right)$ or equivalently, for an invertible matrix $M$,$det\left(M\right)\left({M}^{T}{\right)}^{-1}\left(a×b\right)=Ma×Mb$Remark. The most general formula is$\mathrm{C}\mathrm{o}\left(M\right)\left(a×b\right)=Ma×Mb$where $\mathrm{C}\mathrm{o}\left(M\right)$ is the matrix of cofactors of $M$. Here we do not need $M$ to be invertible. This follows from the proved formula by continuity.