Prove $\frac{\partial (x,y,z)}{\partial (u,v,w)}\frac{\partial (u,v,w)}{\partial (x,y,w)}=1$

Question

If $$x=f(u,v,w)$$ $$y=g(u,v,w)$$ $$z=h(u,v,w)$$

Provided $\frac{\partial(x,y,z)}{\partial(u,v,w)\neq 0}$, prove $$\frac{\partial (x,y,z)}{\partial (u,v,w)}\frac{\partial (u,v,w)}{\partial (x,y,w)}=1$$ I tried to use the property of determinant, if for matrix $$M=AB$$ then $$det(M)=det(A)det(B)$$ So I want to prove that the product of these two Jacobian equals the determinant of the product of these two matrix. I wish the later one would be easier, but I did not go anywhere... Any help? Thanks~

Answer 1

We can write the differential system that describes the transformation from $(u,v,w)$ to $(x,y,z)$ in matrix form as

$$\begin{bmatrix} dx\\dy\\dz \end{bmatrix}=\begin{bmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{bmatrix}\begin{bmatrix} du\\dv\\dw \end{bmatrix} \tag 1$$

and the differential system that describes the inverse transformation as

$$\begin{bmatrix} du\\dv\\dw \end{bmatrix}=\begin{bmatrix} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}&\frac{\partial u}{\partial z}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}&\frac{\partial v}{\partial z}\\\frac{\partial w}{\partial x}&\frac{\partial w}{\partial y}&\frac{\partial w}{\partial z} \end{bmatrix}\begin{bmatrix} dx\\dy\\dz \end{bmatrix} \tag 2$$

Substituting $(1)$ into $(2)$ reveals

$$\begin{bmatrix} dx\\dy\\dz \end{bmatrix}=\begin{bmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}&\frac{\partial u}{\partial z}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}&\frac{\partial v}{\partial z}\\\frac{\partial w}{\partial x}&\frac{\partial w}{\partial y}&\frac{\partial w}{\partial z} \end{bmatrix}\begin{bmatrix} dx\\dy\\dz \end{bmatrix} \tag 3$$

which holds for all $dx$, $dy$, and $dz$. Therefore, the we find that $(3)$ implies

$$\begin{bmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}&\frac{\partial x}{\partial w}\\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}&\frac{\partial y}{\partial w}\\\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}&\frac{\partial z}{\partial w} \end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}&\frac{\partial u}{\partial z}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}&\frac{\partial v}{\partial z}\\\frac{\partial w}{\partial x}&\frac{\partial w}{\partial y}&\frac{\partial w}{\partial z} \end{bmatrix}=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$$

as was to be shown!