Program ( :Solve By Factorization with Pivoting )
思路及原理:
就得到:
程序:
function X=Ni(A) %Input - A is an N x N matrix %Output - I is an N x N inverse matrix of A %and I(j,:)containing the solution to AX(:,j) =E(:,j). %Initialize X, Y,the temporary storage matrix C, and the row % permutation information matrix R [N,N]=size(A); B=eye(N); %B is an N x N identity matrix X=zeros(N,N); Y=zeros(N,N); C=zeros(1,N); R=1:N; %the next steps is to find the factorization(factorize for only once) for p=1:N-1 %Find the pivot row for column p [max1, j]=max(abs(A(p:N,p))); %Interchange row p and j C=A(p,:); A(p,:)=A(j+p-1,:); A(j+p-1,:)=C; d=R(p); R(p)=R(j+p-1); R(j+p-1)=d; if A(p,p)==0 'A is singular. No unique solution' break end %Calculate multiplier and place in subdiagonal portion of A for k=p+1:N mult=A(k,p)/A(p,p); A(k,p) = mult; A(k,p+1:N)=A(k,p+1:N)-mult*A(p,p+1:N); end end for j=1:N %when j is fixed then the method is similar to the Program 3.3 %Solve for Y(:,j) Y(1,j) = B(R(1),j); for k=2:N Y(k,j)= B(R(k),j)-A(k,1:k-1)*Y(1:k-1,j); end %Solve for X(:,j) X(N,j)=Y(N,j)/A(N,N); for k=N-1:-1:1 X(k,j)=(Y(k,j)-A(k,k+1:N)*X(k+1:N,j))/A(k,k); end end
如果运行程序可以看到:Ni(A)和inv(A)运算得到的逆矩阵是相同的
而且 A*Ni(A)=E 所以结果是令人满意的
此方法中LU非直接三角分解只用了一次,通过增加一个j的循环,实现方程组的逐个求解,将得到的N个解向量C(:,j)合到X中得到最终结果。
个人感觉此方法的稳定性不错,暂时不需要改进了吧。
设计这个程序的时候并没遇到什么困难,思路已经想好了:
求解N个方程AXj=Ej;只要通过两个矩阵的对应列向量来存储Xj和Ej以及增加一个j循环,然后利用原来的Program就能达到预期的目的。