Problem with computing inverse using LU

14 visualizaciones (últimos 30 días)
Marc Edwin Montilla
Marc Edwin Montilla el 19 de Ag. de 2014
Comentada: Mariana Pinheiro el 15 de Ag. de 2020
Hi! I seem to have a problem getting the exact inverse of a matrix using LU. This is the code I made, I already have the code for formulating the L and U, this is just a the inverse part for testing.
l = [ 2 0 0 0
-1 1.5 0 0
0 -1 4/3 0
0 0 -1 1.25];
u = [1 -0.5 0 0
0 1 -2/3 0
0 0 1 -0.75
0 0 0 1];
n = length(a);
x = zeros(n,1);
c = zeros(n,1);
d = zeros(n,1);
inverse = zeros(n);
c(1) = 1;
d(1) = c(1) / l(1,1);
for k=1:n
for i=2:n
sum = 0;
for j=1:i-1
sum = sum + l(i,j) * d(j);
end
d(i) = (c(i) - sum) / l(i,i);
end
x(n) = d(n) / u(n,n);
for i=n-1:-1:1
sum = 0;
for j=i+1:n
sum = sum + u(i,j) * x(j);
end
x(i) = [d(i) - sum] / u(i,i);
end
c(k)=0;
c(k+1)=1;
inverse(:,k) = x;
end
This is the result of my code:
inverse =
0.8 1.4 1.2 1
0.6 1.8 1.4 1
0.4 1.2 1.6 1
0.2 0.6 0.8 1
while the true inverse is
0.8 0.6 0.4 0.2
0.6 1.2 0.8 0.4
0.4 0.8 1.2 0.6
0.2 0.4 0.6 0.8
I tested it and I think that the problem may be in the outermost for loop. I just don't know specifically. Thanks in advance!
  2 comentarios
Jutaporn Artniyom
Jutaporn Artniyom el 27 de Abr. de 2020
What is the value of c represent for, and if it's possible to explain how this script work thanks a lot
Mariana Pinheiro
Mariana Pinheiro el 15 de Ag. de 2020
Can you provide the code, please?

Iniciar sesión para comentar.

Respuesta aceptada

Yucheng Ma
Yucheng Ma el 19 de Ag. de 2014
It is my understanding that you would like to implement a C-style matrix inverse procedure using LU decomposition in MATLAB. The code above has a minor mistake in computing the inverse of the L matrix, i.e. "d(1)" is initialized but never updated. I rewrote part of the code and pointed out the difference in the comments. Please refer to the attached file "invLU.m".
In MATLAB, you can use the "inv" function to calculate the inverse of a matrix. You can also use the "mldivide" operator("\") to solve systems of linear equations. The "\" operator is more efficient than explicitly calculating the inverse of a matrix, and can handle singular matrices and sparse matrices.
  2 comentarios
Marc Edwin Montilla
Marc Edwin Montilla el 19 de Ag. de 2014
Thanks for pointing out my error! It is working fine now. We are tasked to solve for the inverse of a matrix by only using the LU decomposition specifically so I guess the inv function is just for checking. I was just wondering when you said that I am implementing a C-style procedure, Is there any other "style"? Anyways, thanks for the suggestions!
Mariana Pinheiro
Mariana Pinheiro el 15 de Ag. de 2020
Can you provide the code, please?

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Linear Algebra en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by