- #1
bananabandana
- 112
- 5
Homework Statement
Don't understand why the inverse jacobian has the form that it does.
Homework Equations
$$ J = \begin{pmatrix} \frac{\partial{x}}{\partial{u}} & \frac{\partial{y}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}} & \frac{\partial{y}}{\partial{v}} \end{pmatrix} $$
$$ J^{-1}= \begin{pmatrix} \frac{\partial{u}}{\partial{x}} & \frac{\partial{v}}{\partial{x}} \\ \frac{\partial{u}}{\partial{y}} & \frac{\partial{v}}{\partial{y}} \end{pmatrix} $$
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! :(
Can somebody please help?
Thank you!