數(shù)值計算求矩陣逆fortran源代碼

1、!此法利用LU分解然后再利用上三角矩陣求逆方法其中利用矩陣乘法以及轉(zhuǎn)置估計精度不敢保證，不過對于低階矩陣，精度可以達到要求。programinverse_matrix!利用上三角矩陣逆矩陣的求法構(gòu)造兩個三角矩陣再分開求逆再求乘積implicitnoneintegeri,nreal*8,allocatable:a(:,:),l(:,:),u(:,:),lt(:,:),inverse_lt(:,:),inverse_l(:,:),inverse_u(:,:),inverse_a(:,:)write(*,*)'PleaseinputtheorderofMatrixA:'read(*,

2、*)nallocate(a(n,n),l(n,n),u(n,n),lt(n,n),inverse_lt(n,n),inverse_l(n,n),inverse_u(n,n),inverse_a(n,n)doi=1,nwrite(*,"('The',i2,'rowofthematrixais:')")iread(*,*)a(i,:)enddocallLU_break(a,n,l,u)callinverse_uptri_matrix(u,n,inverse_u)calltranspose_matrix(l,n,n,lt)callinverse_u

3、ptri_matrix(lt,n,inverse_lt)calltranspose_matrix(inverse_lt,n,n,inverse_l)callmultiply_matrix(inverse_u,n,n,inverse_l,n,inverse_a)doi=1,nwrite(*,"('The',i2,'rowoftheinverseofmatrixais:')")iwrite(*,*)inverse_a(i,:)enddostopendsubroutineLU_break(A,n,L,U)!普通LU分解，目的是為了便于求解逆矩陣im

4、plicitnoneintegerk,i,j,t,nreal*8:A(n,n),L(n,n),U(n,n)real*8sum1,sum2dok=1,ndoj=k,nsum1=0dot=1,k-1sum1=sum1+L(k,t)*U(t,j)enddoU(k,j)=A(k,j)-sum1enddodoi=k+1,nsum2=0dot=1,k-1sum2=sum2+L(i,t)*U(t,k)enddoL(i,k)=(A(i,k)-sum2)/U(k,k)enddoenddodoj=1,nL(j,j)=1enddoreturnendsubroutinetranspose.matrix(A,n,m,B

5、)!求矩陣轉(zhuǎn)置implicitnoneintegeri,j,n,mreal*8A(n,m),B(m,n),tdoi=1,mdoj=1,nB(i,j)=A(j,i)enddoenddoreturnendsubroutinemultiply_matrix(A,n,m,B,l,C)!矩陣間乘法implicitnoneintegeri,j,k,l,n,mreal*8sumreal*8A(n,m),B(m,l),C(n,l)doi=1,ndoj=1,lsum=0dok=1,msum=sum+A(i,k)*B(k,j)enddoC(i,j)=sumenddoenddoreturnendsubroutineinverse_uptri_matrix(A,n,inverse_A)!上三角矩陣的逆矩陣函數(shù)implicitnoneintegeri,j,k,nreal*8sumreal*8A(n,n),inverse_A(n,n)doi=1,ninverse_A(i,i)=1/A(i,i)enddodoi=1,ndoj=1,nif(i<j)thensum=0dok=i