[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