MATLAB EXAMPLES Matrix Solution Methods 58:110 Computer-Aided Engineering Spring 2005 MATLAB EXAMPLES Matrix Solution Methods Department of Mechanical and industrial engineering January 2005

Some useful functions det (A) Determinant lu(A) LU decomposition cond(A) Matrix condition number inv(A) Matrix inverse rank(A) Matrix rank diag, trace Vector and sum of the diagonal elements tril,triu Lower and upper triangular part of matrix cgs, pcg (Preconditioned) conjugate gradients iterative linear equation solver

LU decomposition Here is an example to use LU decomposition for 4X4 matrix A: >> A=[7 3 -1 2;3 8 1 -4; -1 1 4 -1; 2 -4 -1 6] A = 7 3 -1 2 3 8 1 -4 -1 1 4 -1 2 -4 -1 6   >> [l u p]=lu(A) l = 1.0000 0 0 0 0.4286 1.0000 0 0 -0.1429 0.2128 1.0000 0 0.2857 -0.7234 0.0898 1.0000   u = 7.0000 3.0000 -1.0000 2.0000 0 6.7143 1.4286 -4.8571 0 0 3.5532 0.3191 0 0 0 1.8862 p = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 Note: p is a permutation matrix corresponding to the pivoting strategy used. If the matrix is not diagonally dominant, p will not be an identity matrix.

Condition number In MATLAB, Condition number is defined by: To calculate condition number of matrix A by MATLAB built-in function: >> cond(A) ans = 13.7473 Or by the definition, the product of 2-norm of A and inverse A: >> norm(A,2)*norm(inv(A),2)

Iterative methods – Jocobi method The following M-file shows how to use Jocobi method in MATLAB: (jocobi_example.m) if j==i continue else s=s+A(i,j)*x0(j); end x1(i)=(b(i)-s)/A(i,i); if norm(x1-x0)<erp break x0=x1; % show the final solution x=x1 % show the total iteration number n_iteration=k % Iterative Solutions of linear euations:(1) Jocobi Method % Linear system: A x = b % Coefficient matrix A, right-hand side vector b A=[7 3 -1 2; 3 8 1 -4; -1 1 4 -1; 2 -4 -1 6]; b= [-1;0;-3;1]; % Set initial value of x to zero column vector x0=zeros(1,4); % Set Maximum iteration number k_max k_max=1000; % Set the convergence control parameter erp erp=0.0001; % Show the q matrix q=diag(diag(A)) % loop for iterations for k=1:k_max for i=1:4 s=0.0; for j=1:4

Iterative methods- Jocobi method Running M-file in command window: >> jocobi_r_ex q = 7 0 0 0 0 8 0 0 0 0 4 0 0 0 0 6 x = -0.9996 0.9996 -0.9999 0.9996 n_iteration = 60

Iterative methods - Gauss-Seidel method In the M-file for Gauss-seidel method, the only difference is the q matrix, which replaced by: q=tril(A), the codes in the loop changed as the following: for i=1:4 s1=0.0; s2=0.0; if (i==1) continue else for j=1:i-1 s1=s1+A(i,j)*x1(j); end for j=i+1:4 s2=s2+A(i,j)*x0(j); x1(i)=(b(i)-s1-s2)/A(i,i); Running M-file in command window >> gauss_example q = 7 0 0 0 3 8 0 0 -1 1 4 0 2 -4 -1 6 x = -0.9996 0.9997 -0.9999 0.9997 n_iteration = 10

Iterative methods - SOR method Again, In the M-file for SOR method, the only difference is the q matrix, which replaced by: q1=tril(A)-diag(diag(A)); q2=diag(diag(A))/1.4; q=q1+q2; Note: the relaxation factor set to 1.4 for this case. for i=1:4 s1=0.0; s2=0.0; if (i==1) continue else for j=1:i-1 s1=s1+A(i,j)*x1(j); end for j=i+1:4 s2=s2+A(i,j)*x0(j); x1(i)=r*(b(i)-s1-s2)/A(i,i)+(1.0-r)*x0(i); Run this M-file: >> SOR_example q = 5.0000 0 0 0 3.0000 5.7143 0 0 -1.0000 1.0000 2.8571 0 2.0000 -4.0000 -1.0000 4.2857 r = 1.4000 x = -0.9996 0.9997 -0.9999 0.9997 n_iteration = 12

Iterative methods – Conjugate gradient method Use cgs, pcg to solve above linear equations, the tolerance is 1.0E-6 (default) >> x=cgs(A,b,0.00001) cgs converged at iteration 4 to a solution with relative residual 1.2e-015 x = -1.0000 1.0000 >> x=pcg(A,b,0.00001) pcg converged at iteration 4 to a solution with relative residual 3.1e-016

Iterative methods - Summary Solution Total Iteration number Jocobi method ( -0.9996, 0.9996, -0.9999, 0.9996) 60 Gauss –Seidel method (-0.9996, 0.9997, -0.9999, 0.9997) 10 SOR method (-1.0000, 0.9999, -1.0000, 1.0000) 12 Conjugate Gradient method (-1.0000, 1.0000, -1.0000, 1.0000) 4 Note: the exact solution is (-1,1,-1,1)