Finding inv(A) for Ax=b system
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
I tried 3 methods to solve the system for inv(A), where A is highly sparse matrix spy(A) is given below:
1) pinv(A)......... Matlab solve it in 190 sec without any warnings.
2) A\b............. Matlab took only 10 sec but gives warning "Matrix is singular to working precision"
3) [L,U]=lu(A); inv(L)*inv(U) .......... Matlab took 50 sec but give the same warning as in 2nd method.
Is there anyway, I can get inv(A) without warnings.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/166370/image.jpeg)
3 Commenti
Jan
il 27 Lug 2017
Modificato: Jan
il 27 Lug 2017
@Asif Arshid: What is your question? You observed that the slash operator has a different sensitivity to detect near to singular matrices than lu and inv(L)*inv(U). What is the condition number of the matrix? Do you think that slash is to pessimistic or the lu method too sloppy? Or are you surprised by the speed of the slash operator?
Risposte (1)
Star Strider
il 27 Lug 2017
4 Commenti
M.Shaarawy
il 20 Mag 2019
Is there regularized parameterized trust region sub problem (RPTRS) in MATLAB to solve this kind of problem?
Vedere anche
Categorie
Scopri di più su Linear Algebra in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!