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John
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L U decomposition

Asked by John
on 15 Feb 2011
Latest activity Commented on by Daz
on 3 Feb 2015
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

  1 Comment

Daz
on 3 Feb 2015
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

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6 Answers

Answer by Oleg Komarov on 15 Feb 2011

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

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John
Answer by John
on 15 Feb 2011

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.

  1 Comment

Oleg Komarov on 15 Feb 2011
Then can you post the undesired result and the desired one? It's not very clear from your first description.

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Answer by Tan Edwin on 15 Feb 2011

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

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John
Answer by John
on 15 Feb 2011

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

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Answer by Mohamed Said Attia on 4 Jun 2011

*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

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Answer by Mohamed Said Attia on 4 Jun 2011

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

  1 Comment

Walter Roberson
on 4 Jun 2011
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.

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