1.在matlab中用lu分解求逆矩阵的逆矩中检燕窝溯源码可靠吗源代码

在matlab中用lu分解求逆矩阵的源代码

       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