L U decomposition

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

[Daz]

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]

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?

Answer 1

[Oleg Komarov]

Matlab is case-sensitive, if you want to store the output of x then in the first line change X to lowercase.

Oleg

Answer 2

[Mohamed Said Attia]

*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

[ela mti]

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]

undefined function or variable m

[Walter Roberson]

Refer back to the original question; the Answer here only shows the changes instead of copying everything before then as well.

Answer 3

[Mohamed Said Attia]

and when you call the function from matlab use 
[L,U,X]=LU_Parker(A,B) not  LU_Parker(A,B)

[Walter Roberson]

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.

Answer 4

[John]

I tried this but it still outputs my answer the same way, I originally had it as a lowercase x but I changed it to upper case after I realized it didn't change anything.

[Oleg Komarov]

Then can you post the undesired result and the desired one? It's not very clear from your first description.

Answer 5

[Tan Edwin]

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

Answer 6

[John]

Yes, redefining the x like you said allowed the function to output what I was needing, however I must have an error in my coding because I inputed the following matrices and got the following answer but I am getting a 0 for one of the answers which should not be there. Any possible solutions?

INPUT

A=[ 6 0 0 0 0; 0 1 0 -2 0; 1 0 -3 0 0; 0 8 -4 -3 -2; 0 2 0 0 -1];
B=[1;0;0;1;0];
LU_Parker(A,B)

Output:

X =

        0.1667
             0
        0.0432
        0.1841
        1.7778
    ans =
        1.0000         0         0         0         0
             0    1.0000         0         0         0
        0.1667         0    1.0000         0         0
             0    8.0000    1.3333    1.0000         0
             0    2.0000         0    0.3077    1.0000

Answer 7

[Mahesh Prajapati]

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

[Walter Roberson]

Suggestions about what?

One thing I would suggest is that you format the code you post.

[Gudi Vara Prasad]

% 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]

Do you have sample of code in LU Decomposition for a 24x24 matrix using MatLab