Inverse of the Jacobian in the Finite Element Method

Question

In Bathe's Finite Element Procedures 2014 P346, the Jacobian is defined as follows: \begin{equation} \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} \end{equation}

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

So I figured I would try to use the Jacobian inverse to get these values as follows: \begin{equation} \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} \end{equation}

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: \begin{equation} {\mathbf{J}^T}^{-1} \, \mathbf{J}^T \neq \mathbf{I} \end{equation}

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!

Answer 1

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): \begin{equation} \frac{\partial r}{\partial x}, \quad \frac{\partial r}{\partial y}, \quad \frac{\partial s}{\partial x}, \quad \frac{\partial s}{\partial y} \end{equation}

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{equation} \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} \end{equation}

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

Now to proving that reversing the multiplication order of the Jacobian and its inverse yields also the identity matrix (this is my mistake): \begin{equation} \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} = \end{equation}

Component 1,1: \begin{equation} \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 \end{equation}

Component 1,2: \begin{equation} \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 \end{equation}

Component 2,1: \begin{equation} \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 \end{equation}

Component 2,2: \begin{equation} \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 \end{equation}

which proves that the derived inverse Jacobian is correct.