Show that the product of the Jacobian and the inverse Jacobian is 1 I have seen the following fact in a textbook, but am having trouble proving it. If the Jacobian ("stretch factor" for change-of-variables) is given by 
$\left | \frac{\partial (x,y)}{\partial (u,v)} \right | =  \begin{vmatrix}
 \frac{\partial x}{\partial u}&  \frac{\partial y}{\partial u}\\ 
  \frac{\partial x}{\partial v}&   \frac{\partial y}{\partial v}
\end{vmatrix}$, with $x=x(u,v)$ and $y=y(u,v)$,
then 
$\left | \frac{\partial (x,y)}{\partial (u,v)} \right | \left | \frac{\partial (u,v)}{\partial (x,y)} \right | = 1$. 
I am attempting to show this is true by multiplying the matrices and then taking the determinant of the product, but I do not understand how the product becomes equivalent to $\begin{bmatrix}
 1& 0\\ 
  0& 1
\end{bmatrix}$.
Thanks!
 A: We have $$x = x(u,v)  \ \ \ \text{and} \ \ \ y = y(u,v)$$
then the chain rule for functions of several variables states that
$$ \frac{\partial x}{\partial x} = \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x}$$ But $$\frac{\partial x}{\partial x} = 1$$
I'll use these facts later on. For now,

I believe we are trying to prove $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I$$ where $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$
  We'll also prove $$\Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1$$ 

So we begin:
$$\frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix}\begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{vmatrix}$$
$$ = \det \Bigg (\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix} \Bigg ) \det \Bigg (\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
by the identity $$\det(AB) = \det(A) \det(B)$$
so $$ \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{bmatrix}\begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial x}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial x}{\partial v} \frac{\partial v}{\partial y} \\ \frac{\partial y}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial x} & \frac{\partial y}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial y}{\partial v} \frac{\partial v}{\partial y}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & \frac{\partial x}{\partial x}\end{bmatrix} \Bigg )$$
$$ = \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) $$
by the chain rule for functions of several variables, using what we found above.
Now, since $x$ is not a function of $y$, and $y$ is not a function of $x$, we have $$ \frac{\partial x}{\partial y} = 0 \ \ \ \text{and} \ \ \ \frac{\partial y}{\partial x} = 0$$
so $$ \det \Bigg ( \begin{bmatrix} 1 & \frac{\partial x}{\partial y} \\ \frac{\partial y}{\partial x} & 1\end{bmatrix} \Bigg ) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = \det(I)$$
so we've proven that $$ \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = I $$
Now, $$ \det(I) = \det \Bigg (\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} \Bigg ) = 1 $$
so we've also proven that $$ \Bigg \lvert \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\Bigg \rvert = 1 $$ as required
A: Do the calculation straight forward.
$$
\frac{\partial (x,y)}{\partial(u,v)} = 
    \begin{vmatrix}
    x_u & x_v \\
    y_u & y_v \\
    \end{vmatrix}
$$
$$
\frac{\partial (x,y)}{\partial(u,v)} = 
    \begin{vmatrix}
    u_x & u_y \\
    v_x & v_y \\
    \end{vmatrix}
$$
So, we want to prove is,
$$
\frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = x_uy_vu_xv_y + xvy_uu_yv_x - x_uy_vu_yv_x - x_vy_uu_xv_y = 1
$$
Since $x = x(u, v)$, $y = y(u, v)$,
Thus,
$$dx = x_udu + x_vdv$$
$$dy = y_udu + y_vdv$$
By chain rule,
$$\frac{dx}{dx} = x_uu_x + x_vv_x = 1 \qquad (1)$$
$$\frac{dx}{dy} = x_uu_y + x_vv_y = 0 \qquad (2)$$
$$\frac{dy}{dy} = y_uu_y + y_vv_y = 1 \qquad (3)$$
$$\frac{dy}{dx} = y_uu_x + y_vv_x = 0 \qquad (4)$$
Then, showing $$(1) \times (3) - (2) \times (4) = \frac{\partial (x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)} = x_uy_vu_xv_y + xvy_uu_yv_x - x_uy_vu_yv_x - x_vy_uu_xv_y = 1$$
Q.E.D
