How to fix in Gauss formula

Hello,

I have here two formulas of Jacobi and Gauss.

I wrote code in the method Jacobi and it worked for me, but in Gauss there is a problem it came out to me a matrix instead of a vector.

I have a problem with B and c in Gauss, If I did right.

Thanks for the helpers

%Jacobi 
L = tril(A,-1); 
U = triu(A,1);
D = diag(A);
B = -1./D.*(L+U);
c = (1./D).*b;
x_new = B*x+c;

%Gauss 
L = tril(A,-1); 
U = triu(A,1);
D = diag(A);
B = -1./(L+D).*U;
c = (1./(L+D)).*b;
x_new = B*x+c;

Answer 1

[Jon]

It looks like in your calculation of c in both the Jacobi and Gauss you should do a Matrix -vector multiply not an element by element so replace yours with

c = (1./D)*b;

and

c = (1./(L+D))*b;

Maybe there are some other issues too.

Also as a comment you compute the same terms multiple times which isn't too efficient, e.g. 1./(L+D) gets computed twice in your Gauss calculation.

[Jon]

I think you have some additional problems too For example where you have

D = diag(A)

will give D as a vector. But in your formula you take the inverse of this which is not defined for a vector. Do you want a square matrix with the main diagonal of A as the main diagonal and other elements zero? If so you could use

D = diag(diag(A))

[Emilia]

I also tried to do it did not help me.

Look what I got

[Emilia]

[Jon]

Your general problem is that you are thinking that for a matrix M that you can compute it's inverse as 1 . / M . This is not correct. 1 ./ M will invert each element of M individually, which isn't what you want. You should instead use inv or raise the matrix to ^-1

So something like this, where I have left your original code as a comment (please check for sign errors)

%Jacobi 
L = tril(A,-1); 
U = triu(A,1);
D = diag(diag(A));
B = -inv(D)*(L+U);% B = -1./D.*(L+U);
c = inv(D)*b;% c = (1./D).*b;
x_new= B*x+c;


%Gauss 
L = tril(A,-1); 
U = triu(A,1);
D = diag(diag(A));
B = -inv(L+D)*U;% B = -1./(L+D).*U;
c = inv(L+D)*b; % c = (1./(L+D)).*b;
x_new = B*x+c;

[Jon]

As one other fine point. Where you are multiplying both sides of an equation by a matrix A to solve for x, e.g. Ax = b and you solve using x = inv(A)*b it is better, (more numerically efficient and robust) to use MATLABs \ operator so x = A\b

[Emilia]

Excellent it works for me. Thank you :)

[Jon]

Glad to hear it!