By Albrecht Küster (hcak@gmx.de) and Hans Dolhaine (h.dolhaine@gmx.de)
In a foregoing post
https : // community.wolfram.com/groups/-/m/t/1651695?pp auth = EmJqMz34
Stan Gianzero asked, how the inverse of a 2 x 2 matrix, whose elements are 2 x 2 matrices themselves might be found. In that post an answer was given and Stan's problem solved.
But this anwer is highly specialized, because all the elements of the given matrix must be regular ( = non-singular) and moreover certain functions of these elements must be invertible as well. That is generally not the case.
There are more general solutions of Stan's problem, and they might even be expanded to m x m matrices with n x n matrices as elements.
A straightforward (brutal-force) consideration of matrix-products and analysis of matrix-elements yields a "normal" inversion-problem - here in fact the solution of a linear system of equations in the components of the sought inverse, which means there are ( m n )^2 equations to be solved. This method will be called Method 1 in the attached notebook.
There is a considerably simpler and more effective and therefore faster method to find the desired inverse (if it exists) due to the fact, that there are mathematical equivalences shown in the attached paper.
Basically the method can be described by this.
1) Forget the structure of the given matrix m1 to yield matrix m2
2) Find the inverse m3 of m2
3) Reimpose the given structure to m3 to give matrix m4. Then m4 is inverse to m1.
It must be pointed out that in the notebook no means were taken to avoid the case that the inverse might not exist. In this case normal error messages should be produced.
Both methods are shown in the attached notebook and the underlying mathematics are given in the other attachement..
Attachments:
