Prove that product of two matrices is $I$

Question

The pair of variables $(x, y)$ are each functions of the pair of variables $(u, v)$ and vice versa. Consider the matrices: $$ A=\left(\begin{matrix} \frac{\partial{x}}{\partial{u}}& \frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}} & \frac{\partial{y}}{\partial{v}}\end{matrix}\right)\quad\text{ and }\quad B=\left(\begin{matrix} \frac{\partial{u}}{\partial{x}}& \frac{\partial{u}}{\partial{y}} \\ \frac{\partial{v}}{\partial{x}} & \frac{\partial{v}}{\partial{y}}\end{matrix}\right)$$ Prove that $AB = I$, where $I$ is the $2\times 2$ identity matrix.

So, I multiplied these two matrices together, and I got $$AB = \left(\begin{matrix} \frac{\partial{x}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{x}}{\partial{v}}\frac{\partial{v}}{\partial{x}} & \frac{\partial{x}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{x}}{\partial{v}}\frac{\partial{v}}{\partial{y}} \\ \frac{\partial{y}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{v}}{\partial{x}} & \frac{\partial{y}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{v}}{\partial{y}}\end{matrix}\right). $$

Now I have to prove that $AB = \left(\begin{matrix}1 &0\\0&1\end{matrix}\right)$since $x = f(u, v)$ and $y = g(u, v)$, and $u,v$ are some functions of say $s$, by the multivariate chain rule we have: $$ \frac{dx}{ds} = \frac{\partial{x}}{\partial{u}}\frac{du}{ds} +\frac{\partial{x}}{\partial{v}}\frac{dv}{ds}.$$ I thought about substituting $x=s$ (can I do that...?), so that $$1=\frac{dx}{dx} = \frac{\partial{x}}{\partial{u}}\frac{du}{dx} +\frac{\partial{x}}{\partial{v}}\frac{dv}{dx},$$ and from here we get that the upper left element of $AB$ is $1$. Similarly, we could prove that the bottom right element is $1$.

But I have little clue as to how should I prove that $$\frac{\partial{x}}{\partial{u}}\frac{\partial{u}}{\partial{y}}+\frac{\partial{x}}{\partial{v}}\frac{\partial{v}}{\partial{y}} = 0, $$ for instance. Any help? (and while I don't ,,require'' a very rigorous solution, I'm already quite confused by multivariate calculus, so a solution without much handwaving would be appreciated! :) )

Answer 1

$x$ and $y$ are independent variables

$$ \frac{\partial x}{\partial y} = \frac{\partial y}{\partial x} = 0 $$

With this you can repeat the same argument to show the off-diagonal elements are zero

Answer 2

The chain rule is best stated as a formula for infinitesimals rather than either a total or partial derivative; you can work out the correct derivative later. We have $df=\partial_u f du+\partial_v f dv$ (where $\partial_a b$ is an abbreviation for $\frac{\partial_b}{\partial_a}$), so $\partial_x f=\partial_u f\partial_x u+\partial_v f\partial_x v$, which for $f$ equal to $x$ or $y$ gives what you need.