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I need to write a program to solve matrix equations Ax=b where A is an nxn matrix, and b is a vector with n entries using LU decomposition. Unfortunately I'm not allowed to use any prewritten codes in Matlab. I am having problems with the first part of my code where i decompose the matrix in to an upper and lower matrix.

I have written the following code, and cannot work out why it is giving me all zeros on the diagonal in my lower matrix. Any improvements would be greatly appreciated.

function[L R]=LR2(A)



%Decomposition of Matrix AA: A = L R



z=size(A,1);



L=zeros(z,z);



R=zeros(z,z);



for i=1:z



% 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)*R(j,k);



end



L(i,k) = L(i,k)/R(k,k);



end



% Finding R



for k=i:z



R(i,k) = A(i,k);



for j=1:i-1



R(i,k)= R(i,k)-L(i,j)*R(j,k);



end



end



end



R



L



end

I know that i could simply assign all diagonal components of L to be 1, but would like to understand what the problem is with my program!

I am also wondering how to change this program to include pivoting. I understand I need to say that if a diagonal element is equal to zero something needs to be changed. How would I go about this?
Thanks in advance!

For backward and forward elimination I used

% Now use a vector y to solve 'Ly=b' 

 for j=1:z



    for k=1:j-1

       b(j)=b(j)-L(j,k)*b(k);

    end;

    b(j) =b(j)/L(j,j); 

 end;



% Now we use this y to solve Rx = y 

x = zeros( z, 1 );

for i=z:-1:1    

 x(i) = ( b(i) - R(i, :)*x )/R(i, i); 

end

I put that into your code, and it works ;)
For helping you with pivot strategies it would be helpful to know what strategie you want/have to use as there are various versions. One simple version would be to just swap rows such that the diagonal element $a_{ii} neq 0$ for all $i = 1, dots, z$.

function[L R x]=LR2(A,b)

% This program will find a solution to
Ax=b using first giving the
decomposition of the matrix into L and
R and then solving.

% Part 1 - Is this matrix square and
nonsingular? give an error if not

[z y]=size(A);

if (z ~= y )

disp ( 'LR2 error: Matrix must be
square' );

return;

end;

if det(A)==0

disp('L singular error');



return;

end

% Part 2 : Decomposition of matrix
into L and R

L=zeros(z,z);

R=zeros(z,z);

for i=1:z

% 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)*R(j,k);

end

L(i,k) = L(i,k)/R(k,k);

end

% Finding R

for k=i:z

R(i,k) = A(i,k);

for j=1:i-1

R(i,k)= R(i,k)-L(i,j)*R(j,k);

end

end

end

for i=1:z

L(i,i)=1;

end

% Program shows R and L

R

L

% Now use a vector y to solve 'Ly=b'

y=zeros(z,1);

y(1)=b(1)/L(1,1);

for i=2:z

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 Rx = y

x=zeros(z,1);

x(1)=y(1)/R(1,1);

for i=2:z

x(i)=-R(i,1)*x(1);



for k=i:z



    x(i)=x(i)-R(i,k)*x(k);



    x(i)=(y(i)+x(i))/R(i,i);



end;



 end

%print x
x
end

I have solved the original question. However there are still some problems with my solving of Ly=b and Rx=y. I also need to include pivoting. Any improvements would be greatly appreciated again!

After finding the L and U matrix, I tried to find the appropriate B matrix. After then, finding X is a little bit more easier than the traditional way.

% Modifying B

B_new = B;

for i = 2:n

   for k = 1:i-1

       B_new(i,1) = B_new(i,1) - B_new(k,1)*L(i,k); 

   end

end



% Finding X

x = zeros(n, 1);

i = n;

while(i>0)

   x(i) = B_new(i,1);

   k = n;

   while(k > i)

       x(i) = x(i) - x(k)*U(i,k);

       k = k - 1;

   end

   x(i) = x(i) / U(i,i);

   i = i - 1;

end



x

I hope it works for you :)
Greetings from Turkey

NOTE: n equals to z, B equals to b