L U decomposition
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Below I have a code written for solving the L U decomposition of a system of equations however I need my code to just output the answers with this format it outputs the variables in the matrix for example i need the function to output x [1;2;3;4] any suggestions?
function[L,U,X]=LU_Parker(A,B)
[m n]=size(A);
if (m ~= n )
disp ( 'LR2 error: Matrix must be square' );
return;
end;
% Part 2 : Decomposition of matrix into L and U
L=zeros(m,m);
U=zeros(m,m);
for i=1:m
% Finding L
for k=1:i-1
L(i,k)=A(i,k);
for j=1:k-1
L(i,k)= L(i,k)-L(i,j)*U(j,k);
end
L(i,k) = L(i,k)/U(k,k);
end
% Finding U
for k=i:m
U(i,k) = A(i,k);
for j=1:i-1
U(i,k)= U(i,k)-L(i,j)*U(j,k);
end
end
end
for i=1:m
L(i,i)=1;
end
% Program shows U and L
U
L
% Now use a vector y to solve 'Ly=b'
y=zeros(m,1);
y(1)=B(1)/L(1,1);
for i=2:m
y(i)=-L(i,1)*y(1);
for k=2:i-1
y(i)=y(i)-L(i,k)*y(k);
y(i)=(B(i)+y(i))/L(i,i);
end;
end;
% Now we use this y to solve Ux = y
x=zeros(m,1);
x(1)=y(1)/U(1,1);
for i=2:m
x(i)=-U(i,1)*x(1);
for k=i:m
x(i)=x(i)-U(i,k)*x(k);
x(i)=(y(i)+x(i))/U(i,i);
end;
end
2 个评论
Daz
2015-2-3
Aren't you going to get a divide by 0 error? At the very end of what I quoted, you have L(i,k) = L(i,k)/U(k,k);
But the first time through, U is a zero matrix.
L=zeros(m,m); U=zeros(m,m); for i=1:m % Finding L for k=1:i-1 L(i,k)=A(i,k); for j=1:k-1 L(i,k)= L(i,k)-L(i,j)*U(j,k); end L(i,k) = L(i,k)/U(k,k); end
ela mti
2020-11-17
is "i" a counter that shows how many time should loop be done?could you explain that to me?and also "k" and "j" are counter for rows and coluomn?is that so?
回答（7 个）
Oleg Komarov
2011-2-15
1 个投票
Matlab is case-sensitive, if you want to store the output of x then in the first line change X to lowercase.
Oleg
Mohamed Said Attia
2011-6-4
*there is a problem with the way you are solving the equation to get y & x try*
% Now use a vector y to solve 'Ly=b'
y=zeros(m,1); % initiation for y
y(1)=B(1)/L(1,1);
for i=2:m
%y(i)=B(i)-L(i,1)*y(1)-L(i,2)*y(2)-L(i,3)*y(3);
y(i)=-L(i,1)*y(1);
for k=2:i-1
y(i)=y(i)-L(i,k)*y(k);
end;
y(i)=(B(i)+y(i))/L(i,i);
end;
y
% Now we use this y to solve Ux = y
x=zeros(m,1);
x(m)=y(m)/U(m,m);
i=m-1;
q=0;
while (i~= 0)
x(i)=-U(i,m)*x(m);
q=i+1;
while (q~=m)
x(i)=x(i)-U(i,q)*x(q);
q=q+1;
end;
x(i)=(y(i)+x(i))/U(i,i);
i=i-1;
end;
x
3 个评论
ela mti
2020-11-17
would you explain to me this part and what is q ?
i=m-1;
q=0;
while (i~= 0)
x(i)=-U(i,m)*x(m);
q=i+1;
while (q~=m)
x(i)=x(i)-U(i,q)*x(q);
q=q+1;
end;
x(i)=(y(i)+x(i))/U(i,i);
i=i-1;
end;
x
THEOPHILUS GODFREEY
2021-5-12
undefined function or variable m
Walter Roberson
2021-5-12
Refer back to the original question; the Answer here only shows the changes instead of copying everything before then as well.
Mohamed Said Attia
2011-6-4
and when you call the function from matlab use
[L,U,X]=LU_Parker(A,B) not LU_Parker(A,B)
1 个评论
Walter Roberson
2011-6-4
Not really relevant: if you do not specify output variables and do not put a semi-colon at the end of the line, you will get
ans =
for each of the output variables, in left-to-right order.
John
2011-2-15
0 个投票
1 个评论
Oleg Komarov
2011-2-15
Then can you post the undesired result and the desired one? It's not very clear from your first description.
Tan Edwin
2011-2-15
0 个投票
Maybe u can try adding X=x to allow it to ouput the values of x?
not sure if this is what u want.
edwin
Mahesh Prajapati
2020-9-21
编辑：Walter Roberson
2020-12-30
any suggestions?
function[L,U,X]=LU_Parker(A,B)
[m n]=size(A);
if (m ~= n ) disp ( 'LR2 error: Matrix must be square' ); return; end;
% Part 2 : Decomposition of matrix into L and U
L=zeros(m,m);
U=zeros(m,m);
for i=1:m
% Finding L
for k=1:i-1
L(i,k)=A(i,k);
for j=1:k-1
L(i,k)= L(i,k)-L(i,j)*U(j,k);
end
L(i,k) = L(i,k)/U(k,k);
end
% Finding U
for k=i:m
U(i,k) = A(i,k);
for j=1:i-1
U(i,k)= U(i,k)-L(i,j)*U(j,k);
end
end
end
for i=1:m
L(i,i)=1;
end
% Program shows U and L U L
% Now use a vector y to solve 'Ly=b'
y=zeros(m,1);
y(1)=B(1)/L(1,1);
for i=2:m
y(i)=-L(i,1)*y(1);
for k=2:i-1
y(i)=y(i)-L(i,k)*y(k);
y(i)=(B(i)+y(i))/L(i,i);
end;
end;
% Now we use this y to solve Ux = y
x=zeros(m,1);
x(1)=y(1)/U(1,1);
for i=2:m
x(i)=-U(i,1)*x(1);
for k=i:m
x(i)=x(i)-U(i,k)*x(k);
x(i)=(y(i)+x(i))/U(i,i);
end;
end
3 个评论
Walter Roberson
2020-12-30
Suggestions about what?
One thing I would suggest is that you format the code you post.
Gudi Vara Prasad
2021-4-20
编辑：Gudi Vara Prasad
2021-4-20
% There is some mistake with the Back Substituion at the end in the above code. Please check it again..
clc;
clear all;
close all;
format 'short';
% input:
% A = coefficient matrix
% B = right hand side vector
% output:
% x = solution vector
disp('Application of LU Decomposition')
tic
A = % user input
B = % user input
[m, n] = size(A);
if m ~= n, error('Matrix A must be square'); end
% Decomposition of matrix into L and U :
L=zeros(m,m);
U=zeros(m,m);
for i=1:m
% Finding L
for k=1:i-1
L(i,k)=A(i,k);
for j=1:k-1
L(i,k)= L(i,k)-L(i,j)*U(j,k);
end
L(i,k) = L(i,k)/U(k,k);
end
% Finding U
for k=i:m
U(i,k) = A(i,k);
for j=1:i-1
U(i,k)= U(i,k)-L(i,j)*U(j,k);
end
end
end
for i=1:m
L(i,i)=1;
end
% Program shows U and L U L :
% Now use a vector y to solve 'Ly=b' :
y=zeros(m,1);
y(1)=B(1)/L(1,1);
for i=2:m
y(i)=-L(i,1)*y(1);
for k=2:i-1
y(i)=y(i)-L(i,k)*y(k);
y(i)=(B(i)+y(i))/L(i,i);
end
end
fprintf("Lower Decomposition Traingle = ")
L
fprintf("Upper Decomposition Traingle = ")
U
% Now we use this y to solve Ux = y
% x=zeros(m,1);
% x(1)=y(1)/U(1,1);
% for i=2:m
% x(i)=-U(i,1)*x(1);
% for k=i:m
% x(i)=x(i)-U(i,k)*x(k);
% x(i)=(y(i)+x(i))/U(i,i);
% end
% end
% Back substitution :
x = zeros(n, 1);
AM = [U B];
x(n) = AM(n, n+1) / AM(n, n);
for i = n - 1: - 1:1
x(i) = (AM(i, n+1) - AM(i, i + 1:n) * x(i + 1:n)) / AM(i, i);
end
fprintf("Solution of the system is = ")
x
Edmar
2024-2-25
Do you have sample of code in LU Decomposition for a 24x24 matrix using MatLab
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