Solve linear equation in matrix form
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I need to find pi^(0) from this linear equation
pi^(0) is a row vector, e is column vector of 1
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1376309/image.jpeg)
where A_0, R, B, A, and C are square matrix.
Matrix A_0, B, A, and C are given.
R can be found by using successive substitutions method
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1376314/image.png)
After i found R, i use this code to find pi^(0) (q is the same as pi^(0))
q=sym('q',[1 mm]) %row vector
I=eye(mm);
e=ones(mm,1); %column vector of 1
a=q*(inv(I-R))*e %size 1 1
aa=a-1
j=solve(aa,sym('q1')) %j is the same as q1=
w=q*(A_0+R*B) %size 1 row mm column
k=subs(w,sym('q1'),j) %subs q1 with j
[C,S]=equationsToMatrix(k,q)
rank(S)
rank(C) %if rank(C)=rank([C,S]) the system can be solved
rank([C,S])
ji=C\S %ji is pi^(0) which i need to find and must fulfill all value less than 1 and non negative.
But, after i run, rank(C)/=rank([C,S]) so the system can't be solved
Is there any code that is wrong? or there is code that more simple than this code?
0 commentaires
Réponses (1)
Torsten
le 6 Mai 2023
Modifié(e) : Torsten
le 6 Mai 2023
Why don't you use
pi0 = null((A0+R*B).');
if ~isempty(pi0)
pi0 = pi0.'/(pi0.'*inv(I-R)*e)
end
9 commentaires
Torsten
le 12 Mai 2023
Modifié(e) : Torsten
le 12 Mai 2023
I don't understand why you experiment above with solutions to 3 equations out of the 4.
For that your problem has a solution, there must exist a vector different from the null vector that satisfies pi^0*(A_0+R*B) = 0 or (A_0+R*B).'*pi^0.' = 0. If such vector(s) exist, you'll get a basis of the vector space spanned by these vectors by the command null((A_0+R*B).'). If the result of this command is empty, your problem has no solution.
Equivalently you can check whether rank((A_0+R*B).') < 3. If this is the case, you'll also be able to solve your problem.
Voir également
Catégories
En savoir plus sur Operating on Diagonal Matrices dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!