solve a symbolic matrix equation

16 views (last 30 days)
marcel hendrix
marcel hendrix on 1 Sep 2018
Commented: marcel hendrix on 22 Oct 2022
I wonder if it is at all possible to solve the following equation with the Symbolic Toolbox. Matlab was working for more than 5 hours on this when I terminated the process.
The problem could be that the matrices Aa and Ya are not square. Actually I don't need a symbolic answer, numeric would be fine. The main reason to try the Symbolic Toolbox is that I don't see how to approach 'Ya*D' = Aa*Ya', as Ya is being both post- and pre-multiplied. I guess fine points of vector-matrix algebra are needed here.
-marcel
%%Differentiation script 3
syms e Aa Ya D
syms e1 e2 e3 e4
syms a11 a12 a13 a21 a22 a23 a31 a32 a33
syms y11 y12 y13 y14 y21 y22 y23 y24 y31 y32 y33 y34
syms d11 d12 d13 d14 d21 d22 d23 d24 d31 d32 d33 d34 d41 d42 d43 d44
e = [e1 e2 e2 e4];
Aa = [a11 a12 a13
a21 a22 a23
a31 a32 a33];
Ya = [y11 y12 y13 y14
y21 y22 y23 y24
y31 y32 y33 y34];
D = [d11 d12 d13 d14
d21 d22 d23 d24
d31 d32 d33 d34
d41 d42 d43 d44];
solve(Ya*D'- Aa*Ya==0, Ya)
%EOF
  9 Comments
marcel hendrix
marcel hendrix on 22 Oct 2022
Hi David,
It is very likely that you are right. The problem is from circuit simulation, and it assumes the (possibly non-linear) network has a '(periodic) steady-state'. Even when non-periodic, there can be multiple (or no) operating points.

Sign in to comment.

Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!