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Title: Gaussian Elimination Method with Pivoting for Solving a Linear System
Author: Cheuk Lau
Date: 2/5/2014
Description: This function solves a linear system Ax=b using the Gaussian
elimination method with pivoting. The algorithm is outlined below:
1) Initialize a permutation vector r = [1, 2,...,n] where r(i)
corresponds to row i in A.
2) For k = 1,...,n-1 find the largest (in absolute value) element among
a(r(k),k),a(r(k+1),k),...,a(r(n),k).
3) Assume r(j,k) is the largest element. Switch r(j) and r(k).
4) For i=1,...,k-1,k+1,...,n calculate:
zeta = a(r(i),k) / a(r(k),k)
5) For j=k,...,n calculate:
a(r(i),j)=a(r(i),j)-a(r(k),j)*zeta;
b(r(i)) = b(r(i))-b(r(k))*zeta
6) Steps 1 through 6 has effectively diagonalized A.
7) Each element in the solution vector is:
x(r(i)) = b(i)/a(i,i);
Files required: (1) GAUSS_ELIM.m
Reference: (1) http://mathworld.wolfram.com/GaussianElimination.html
Input parameters: A = n by n matrix in standard Matlab notation
b = n by 1 right-hand vector
Output: x = computed solution
Sample input:
% We will set A as the Hilbert matrix:
% A(i, j) = 1 / (i + j - 1)
%
% With the exact solution:
% x(i) = 1
%
% We can calculate the right-hand vector:
% b = Ax
%
% By defining the exact solution we can then calculate a rate of
% error convergence. This method is known as the method of
% manufactured solutions (MMS).
% Assemble A-matrix
n=10; % You can change the dimension size if you want to look at
% error convergence with input size
A = zeros(n);
for i = 1 : 1 : n
for j = 1 : 1 : n
A(i, j) = 1 / (i + j - 1);
end
end
% Exact solution -- used to calculate the error, if desired
x_exact = ones(n, 1);
% Assemble b-vector
b = A*x_exact;
Output:
x = [1.0000
1.0000
1.0000

1.0000
0.9999
1.0002
0.9997
1.0003
0.9998
1.0000]