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]
