[matlab]用LU分解法求解线性方程组

求解Ax=d的解x

1. LU分解

% file: myLU.m

function [L,U]=myLU(A)
%实现对矩阵A的LU分解，L为下三角矩阵
A
[n,n]=size(A);
L=zeros(n,n);
U=zeros(n,n);
for i=1:n
    L(i,i)=1;
end
for k=1:n
    for j=k:n
        U(k,j)=A(k,j)-sum(L(k,1:k-1).*U(1:k-1,j)');
    end
    for i=k+1:n
        L(i,k)=(A(i,k)-sum(L(i,1:k-1).*U(1:k-1,k)'))/U(k,k);
    end
end

2. 用分解获得的L，U求解x

 

[L,U] = myLU(A)

[n,m]=size(A)

y(1)=d(1);
for i=2:n    
    for j=1:i-1
        d(i)=d(i)-L(i,j)*y(j);
    end
    y(i)=d(i);
end

x(n)=y(n)/U(n,n);
for i=(n-1):-1:1
    for j=n:-1:i+1
        y(i)=y(i)-U(i,j)*x(j);
    end
    x(i)=y(i)/U(i,i);
end

　　

转载于:https://www.cnblogs.com/wangshide/archive/2011/12/09/2282641.html