Gauss-Seidel method Help
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Hello,
I am hoping someone can assist me with this problem. I created a Gauss-Seidel code that will allow me to solve a set of linear equations, finding x1, x2 x3 and x4. I made two matrices; A=[4 -21 -7 1; -4 0 -3 11 ; 4 -1 10 -1; 151/8 5 8 -3] and b =[11; 15; 19; -12;]. Through using the equatio Ax=b i would be able to find the unknows, which worked using the backslash built in solver. But my code below give 4 values for x but they do not match the reuslts of the built in solver and do not equate when substitued back into a line of the linear equations.
A major aspect of the code is that it is meant to make your matrix diagonally dominant to solve.
My code is as follows:
function gauss-seidel
A=input('write matrix a')
b=input('write matrix b')
x=linspace(0,0,length(A))';
n=size(x,1);
normVal=Inf;
nmax=1000; %number of maximum iterations which can be reached%
tol=0.000001; % Tolerence for method%
iter=0;
while normVal>tol && iter<nmax
x_old=x;
[V,A]=eig(A);
TF=isdiag(A);
if TF==0
fprintf('the matrix is not diagonally dominant')
else
for i=1:n
guess=0;
for j=1:i-1
guess=guess+A(i,j)*x(j);
end
for j=i+1:n
guess=guess+A(i,j)*x_old(j);
end
x(i)=(b(i)+guess)/(A(i,i));
end
iter=iter+1;
normVal=norm(x_old-x);
end
end
fprintf('Solution of the system is : ')
for i=1:length(x)
fprintf(' %1.4f ',x(i));
end
fprintf('in %d iterations ',iter)
4 个评论
John D'Errico
2020-4-22
Please stop asking the same question day after day.
kemgi
2020-4-23
John D'Errico
2021-11-11
If you made a mistake, it is trivial to edit your question to fix it. But many people seem to post the same question as many as 5 or 6 times. If they don't get an answer quickly enough to suit their needs, they just keep posting the same question.
Mahmoud
2023-3-7
Writing the negative exponent to give the same result as the calculator. Example: 0.3000 * 10^-3 gives a result of 3 * 10^-4
But in MATLAB it gives a very different result. What is the problem here or how to write the equation correctly to give the same value as the calculator
采纳的回答
Prabhan Purwar
2020-4-22
Hi,
Refer to the following code:
The following code performs Gauss-Seidel...
clc
clear
close all
A=[5 -2 3 0 6; -3 9 1 -2 7.4; 2 -1 -7 1 6.7; 4 3 -5 7 9; 2 3.5 6.1 -4 -8.1];
b=[-1 2 3 0.5 3.1]';
x=linspace(0,0,length(A))';
n=size(x,1);
normVal=Inf;
nmax=1000; %number of maximum iterations which can be reached%
tol=1e-3; % Tolerence for method%
iter=0;
while normVal>tol && iter<nmax
x_old=x;
for i=1:n
guess=0;
for j=1:i-1
guess=guess+A(i,j)*x(j);
end
for j=i+1:n
guess=guess+A(i,j)*x_old(j);
end
x(i)=(1/A(i,i))*(b(i)-guess);
end
iter=iter+1;
normVal=norm(x_old-x);
end
fprintf('Solution of the system is : \n%f\n%f\n%f\n%f\n%f in %d iterations',x,iter);
Output:
0.551479
0.469276
-0.595161
-0.649082
-0.171448 in 86 iterations
Tips:
The convergence properties of the Gauss-Seidel method are dependent on the matrix A. Namely, the procedure is known to converge if either:
- A is symmetric positive-definite
https://www.mathworks.com/help/matlab/ref/chol.html (chol, only for symmetric matrix) [~,p] = chol(A)
OR
- A is strictly or irreducibly diagonally dominant
(Diagonal Dominant)
The Gauss-Seidel method sometimes converges even if these conditions are not satisfied.
Although you can prove that the methods converge if A has all eigenvalues positive from the above conditions (but not the only convergence criterion).
Hope it helps!!
3 个评论
kemgi
2020-4-23
Prabhan Purwar
2020-4-23
Kindly accept the answer if its helpful...
ahmed al-khazraji
2021-2-2
thanks for the hard work
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