# Inverse Jacobian

1. Jan 22, 2015

### bananabandana

1. The problem statement, all variables and given/known data
Don't understand why the inverse jacobian has the form that it does.

2. Relevant equations
$$J = \begin{pmatrix} \frac{\partial{x}}{\partial{u}} & \frac{\partial{y}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}} & \frac{\partial{y}}{\partial{v}} \end{pmatrix}$$

$$J^{-1}= \begin{pmatrix} \frac{\partial{u}}{\partial{x}} & \frac{\partial{v}}{\partial{x}} \\ \frac{\partial{u}}{\partial{y}} & \frac{\partial{v}}{\partial{y}} \end{pmatrix}$$

3. The attempt at a solution
This flow of logic makes sense to me:
$$\begin{pmatrix} \frac{\partial{f}}{\partial{u}} \\ \frac{\partial{f}}{\partial{v}} \end{pmatrix} = J \begin{pmatrix} \frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}} \end{pmatrix}$$
(due to the equation for change of variables in two dimensions).

Similarly, looking at the inverse transformation, if we say that there is some matrix $M$ that represents the inverse of $J$, we know:

$$\begin{pmatrix} \frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}} \end{pmatrix} =M \begin{pmatrix} \frac{\partial{f}}{\partial{u}} \\ \frac{\partial{f}}{\partial{v}} \end{pmatrix}$$

Using the equation of the change of variables in 2D it's clear that $J^{-1}$ should look as shown.

I'm confused however, because if I apply the general equation for the inverse of a $2X2$ matrix to $J$, we get a matrix, $j^{-1}$, like this:

$$j^{-1} = \frac{1}{|J|} \begin{pmatrix} \frac{\partial{y}}{\partial{v}} & - \frac{\partial{y}}{\partial{u}} \\ -\frac{\partial{x}}{\partial{v}} & \frac{\partial{x}}{\partial{u}} \end{pmatrix}$$

But this would then imply that
$$\frac{1}{|J|} \frac{\partial{y}}{\partial{v}} = \frac{\partial{u}}{\partial{x}}$$ etc .. .
Which wouldn't seem to be generally true.

Similarly, if we multiply $J$ and $J^{-1}$, using the standard rules for matrix multiplication, we get:

$$JJ^{-1} = \begin{pmatrix} \frac{\partial{x}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{y}}{ \partial{u}}\frac{\partial{u}}{\partial{y}} & \frac{\partial{x}}{\partial{u}}\frac{\partial{v}}{\partial{x}}+\frac{\partial{y}}{\partial{u}}\frac{\partial{v}}{\partial{y}} \\ \frac{\partial{x}}{\partial{v}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{u}}{\partial{y}} & \frac{\partial{x}}{\partial{v}}\frac{\partial{v}}{\partial{x}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{v}}{\partial{y}} \end{pmatrix}$$

If the above $'J^{-1}$ is a true inverse of $J$ then clearly: $JJ^{-1}=I$, where $I$ is the identity matrix. But I don't understand how the terms in the above expression equate to the corresponding terms in the identity matrix! :(

Thank you!

2. Jan 27, 2015