# Gauss-Seidel iterative method with no relaxation factor

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Dai Nguyen on 27 Oct 2020
Commented: Dai Nguyen on 9 Nov 2020
Hi I want to create a Matlab code for Gauss-Seidel iterative method with no relaxation factor and use the solution using a termination tolerance=.01% for the relative approximate error. This is what I got so far. For some reason my stigma is zero and it won't able to calculate it
A= [60 -40 0; -40 60 -20; 0 -20 20]
b=[29.4; 39.2; 58.8]
x=[0 0 0 0]'
n=size(x,1);
normVal=Inf;
%%
% * _*Tolerence for method*_
tol=1e-5; itr=0;
%%
while normVal>tol
x_old=x;
for i=1:n
sigma=0;
for j=1:i-1
sigma=sigma+A(i,j)*x(j);
end
for j=i+1:n
sigma=sigma+A(i,j)*x_old(j);
end
x(i)=(1/A(i,i))*(b(i)-sigma);
end
itr=itr+1;
normVal=norm(x_old-x);
end
%%
fprintf('Solution of the system is : \n%f\n%f\n%f\n%f in %d iterations',x,itr);

Sreeranj Jayadevan on 9 Nov 2020
I have executed the given code in MATLAB and it is giving me the required output. It seems to me that you have accidentally initialized the solution vector "x" as a 4 by 1 array. Since only three equations are involved in the problem, "x" should be a 3 by 1 array i.e
x=[0 0 0]';
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Dai Nguyen on 9 Nov 2020
Thank you

R2020a

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