# Inverse of the Jacobian in the Finite Element Method

In Bathe's Finite Element Procedures 2014 P346, the Jacobian is defined as follows: $$$$\mathbf{J} = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} \end{bmatrix}$$$$

I need to calculate the following components: $$$$\frac{\partial r}{\partial x}, \quad \frac{\partial r}{\partial y}, \quad \frac{\partial s}{\partial x}, \quad \frac{\partial s}{\partial y}$$$$

So I figured I would try to use the Jacobian inverse to get these values as follows: $$$$\mathbf{J}^T \, {\mathbf{J}^T}^{-1} = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial s} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial s} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \end{bmatrix} = \mathbf{I}$$$$

However, as I understand, the order of matrices multiplication is not important in this particular case. So if I tried to reverse the order of the matrices multiplication and do the math, this wouldn't work! As in: $$$${\mathbf{J}^T}^{-1} \, \mathbf{J}^T \neq \mathbf{I}$$$$

I am not sure what I did wrong here!

PS: The same applies for this document I found online on slide 4. Although the Jacobian is defined differently, they follow the same procedure. However, if you reverse the order of matrices multiplication it's no longer the unity matrix!

• For any invertible matrix, you have that $AA^{-1}=I=A^{-1}A$. If you get something different, you made a mistake somewhere :-) Dec 14, 2021 at 15:57
• I agree something is missing here, I didn't find any information in major books on finite element, that's why I am skeptical about this procedure! Do you know of any other way to calculate the components I need above? Thank you. Dec 14, 2021 at 16:00
• You are trying to form the inverse of the Jacobian from the derivatives of the inverse functions, which should work if you are implementing things correctly. I don't know what code you are using, but it should have a general method to compute the inverse of a matrix, which you can compare with what you are doing. Dec 14, 2021 at 16:04
• I am sorry for wasting your time. I realized I made a mistake in multivariate calculus. I will post the detailed answer shortly. Dec 14, 2021 at 16:26
• That's perfect. Your mistake is probably a mistake that people have fallen for in the past, and will in the future. The answer might help :) Dec 14, 2021 at 16:29

I realized I made a mistake in the question by claiming that the reverse multiplication order does not yield the same result. The correct answer is as follows.

My main quest is to find the following components (needed for calculating the second derivatives of shape functions): $$$$\frac{\partial r}{\partial x}, \quad \frac{\partial r}{\partial y}, \quad \frac{\partial s}{\partial x}, \quad \frac{\partial s}{\partial y}$$$$

Note that due to the multivariate nature of the variables $$x$$ and $$y$$, $$\frac{\partial r}{\partial x} \neq \frac{\partial x}{\partial r}$$. One way (and probably the only way!) to do this is by inverting the Jacobian as follows: $$$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$$$

By solving these four equations in four unknowns, we get that the inverse Jacobian is equivalent to: $$$$\begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial s}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial s}{\partial y} \end{bmatrix}$$$$

Now to proving that reversing the multiplication order of the Jacobian and its inverse yields also the identity matrix (this is my mistake): $$$$\begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} \end{bmatrix} \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial s}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial s}{\partial y} \end{bmatrix} =$$$$

Component 1,1: $$$$\frac{\partial x}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial r}{\partial y} = \frac{\partial r}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial r}{\partial y} \frac{\partial y}{\partial r} = \frac{dr}{dr} = 1$$$$

Component 1,2: $$$$\frac{\partial x}{\partial r} \frac{\partial s}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial s}{\partial y} = \frac{\partial s}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial s}{\partial y} \frac{\partial y}{\partial r} = \frac{ds}{dr} = 0$$$$

Component 2,1: $$$$\frac{\partial x}{\partial s} \frac{\partial r}{\partial x} + \frac{\partial y}{\partial s} \frac{\partial r}{\partial y} = \frac{\partial r}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial r}{\partial y} \frac{\partial y}{\partial s} = \frac{dr}{ds} = 0$$$$

Component 2,2: $$$$\frac{\partial x}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial y}{\partial s} \frac{\partial s}{\partial y} = \frac{\partial s}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial s}{\partial y} \frac{\partial y}{\partial s} = \frac{ds}{ds} = 1$$$$

which proves that the derived inverse Jacobian is correct.

• To even improve your answer you might add how the inverse, adjoint and determinant of a matrix are connected. You may also link terms like covariant and contravariant metrics known in the tensor analysis. Dec 14, 2021 at 18:26