Building up a Matrix using for LOOP and summation

1 visualización (últimos 30 días)
Abdulrahman Odhah
Abdulrahman Odhah el 5 de Abr. de 2020
Comentada: Abdulrahman Odhah el 16 de Abr. de 2020
Hello fellows,
I am trying to make this equation but I couldn't do the summaiton part , i tried to use the double for loop but still not sure how to do the summation part,
I have the matrix " a " already defined 4x4 also initial l and u matrices are defiend l=eye(N),u=zeros(N,N);
any hints?

Respuesta aceptada

David Hill
David Hill el 5 de Abr. de 2020
I believe the below is correct.
I=eye(4);
u=zeros(4);
u(1,1:4)=a(1,1:4);
I(1:4,1)=a(1:4,1);
for i=2:4
for j=1:4
u(i,j)=a(i,j)-sum(I(i,1:i-1).*u(1:i-1,j)');
if i<j
I(j,i)=(I(j,i)-sum(I(j,1:i-1).*u(1:i-1,i)'))/u(i,i);
end
end
end
  2 comentarios
Abdulrahman Odhah
Abdulrahman Odhah el 8 de Abr. de 2020
I think its the correct one also, I'll try it on some values,
thanks a lot, for your suuport,
Abdulrahman Odhah
Abdulrahman Odhah el 16 de Abr. de 2020
I've applied the code on some known values, this code is supposed to find the lu matrices of a matrix A, using the above expression
what i've got from the code given by you is the following,
I =
6.0000 0 0 0
4.0000 1.0000 0 0
-2.0000 -0.5333 1.0000 0
3.0000 0.8000 0.0845 1.0000
u =
6.0000 4.0000 2.0000 1.0000
-20.0000 -15.0000 -8.0000 -2.0000
-0.6667 2.0000 4.7333 8.9333
1.0563 1.8310 4.0000 -1.1549
I thing that the code representing the mathematical expression has to produce the following
I =
1.0000 0 0 0
0.6667 -0.5000 0.5000 1.0000
-0.3333 1.0000 0 0
0.5000 0 1.0000 0
u=
6.0000 4.0000 2.0000 1.0000
0 3.3333 5.6667 8.3333
0 0 3.0000 0.5000
0 0 0 5.2500
Thanks again for your suuport.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Etiquetas

Productos


Versión

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by