The Question
Hi all, I am trying to find the Jacobian matrix of a differential equation that has the following format:
$$\mathbf{A}(\mathbf{y})\dot{\mathbf{y}} = \mathbf{b}(\mathbf{y}) \tag{1}$$
where $\mathbf{A}$ is a square matrix as a function of $\mathbf{y}$, and $\mathbf{b}$ is a vector with the same dimensions as $\mathbf{y}$ which is also a function of $\mathbf{y}$. Ideally I would be able to find the Jacobian of the following:
$$\dot{\mathbf{y}} = \mathbf{A}^{-1}\mathbf{b} \tag{2}$$
however it is extremely hard and laborious to symbolically invert $\mathbf{A}$ since its a $5\times5$ matrix. I was therefore wondering wheather there is a better approach in obtaining the Jacobian matrix of the above, rather than brute-forcing it by taking the inverse.
My Thoughts
I know that we can have a property such $\frac{d\mathbf{A}^{-1}}{dt}=-\mathbf{A}^{-1}\frac{d\mathbf{A}}{dt}\mathbf{A}^{-1}$, which seems to have potential in helping. I notice however in this case $t$ is defined as a scalar and not as a vector as it should be in my case.
Furthermore there are other questions/resources that address similar issues such as this and this, however I can't see how I can extend those to this case that I am dealing with.