How do I found A-B*x which these matrix are 3x3

1 vue (au cours des 30 derniers jours)
Rommel Brito
Rommel Brito le 9 Sep 2020
A=[1231200 -615600 0; -615600 1231200 -615600; 0 -615600 615600]
A =
1231200 -615600 0
-615600 1231200 -615600
0 -615600 615600
B=[62.68 0 0; 0 62.68 0; 0 0 62.68]
B =
62.6800 0 0
0 62.6800 0
0 0 62.6800
And I want to get the A-Bx which x is a uknown variable that I want to know.

Réponses (2)

Walter Roberson
Walter Roberson le 9 Sep 2020
>> x=B\A
x =
19642.6292278239 -9821.31461391193 0
-9821.31461391193 19642.6292278239 -9821.31461391193
0 -9821.31461391193 9821.31461391193
>> A-B*x
ans =
0 0 0
0 0 0
0 0 0
  3 commentaires
Walter Roberson
Walter Roberson le 10 Sep 2020
A = [1231200, -615600, 0; -615600, 1231200, -615600; 0, -615600, 615600];
B = [62.68 0 0; 0 62.68 0; 0 0 62.68];
syms L
vals = solve(det(A - B*L), 'MaxDegree', 3);
real(vpa(vals)) %imaginary component is noise
Rommel Brito
Rommel Brito le 10 Sep 2020
Thank you! That worked!

Connectez-vous pour commenter.


Bruno Luong
Bruno Luong le 11 Sep 2020
Waoh, revise the theory of eigen vectors and linear algebra in general
Multiply a vector by a constant diagonal matrix is like multiply by the scalar.
This equation admits a solution for only when B = lambda*Identity where lambda is eigen value and x is the corresponding eigen eigen vector.
I'm surprises no one tell you that.

Catégories

En savoir plus sur Linear Algebra 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!

Translated by