用c语言编程，求解逆矩阵

2023-02-23 06:56:01Python011

用c语言编程，求解逆矩阵,第1张

#include <stdio.h>#include <stdlib.h>#include <malloc.h>void MatrixOpp(double *A, int m, int n, double* invmat)void MatrixInver(double *A, int m, int n, double* invmat)double Surplus(double A[], int m, int n)int matrix_inv(double* p, int num, double* invmat)void MatrixOpp(double A[], int m, int n, double* invmat){ int i, j, x, y, k double *SP = NULL, *AB = NULL, *B = NULL, X SP = (double *) malloc(m * n * sizeof(double)) AB = (double *) malloc(m * n * sizeof(double)) B = (double *) malloc(m * n * sizeof(double)) X = Surplus(A, m, n) X = 1 / X for (i = 0i <mi++) for (j = 0j <nj++) { for (k = 0k <m * nk++) B[k] = A[k] { for (x = 0x <nx++) B[i * n + x] = 0 for (y = 0y <my++) B[m * y + j] = 0 B[i * n + j] = 1 SP[i * n + j] = Surplus(B, m, n) AB[i * n + j] = X * SP[i * n + j] } } MatrixInver(AB, m, n, invmat) free(SP) free(AB) free(B)}void MatrixInver(double A[], int m, int n, double* invmat){ int i, j double *B = invmat for (i = 0i <ni++) for (j = 0j <mj++) B[i * m + j] = A[j * n + i]}double Surplus(double A[], int m, int n){ int i, j, k, p, r double X, temp = 1, temp1 = 1, s = 0, s1 = 0 if (n == 2) { for (i = 0i <mi++) for (j = 0j <nj++) if ((i + j) % 2) temp1 *= A[i * n + j] else temp *= A[i * n + j] X = temp - temp1 } else { for (k = 0k <nk++) { for (i = 0, j = ki <m, j <ni++, j++) temp *= A[i * n + j] if (m - i) { for (p = m - i, r = m - 1p >0p--, r--) temp *= A[r * n + p - 1] } s += temp temp = 1 } for (k = n - 1k >= 0k--) { for (i = 0, j = ki <m, j >= 0i++, j--) temp1 *= A[i * n + j] if (m - i) { for (p = m - 1, r = ir <mp--, r++) temp1 *= A[r * n + p] } s1 += temp1 temp1 = 1 } X = s - s1 } return X}int matrix_inv(double* p, int num, double* invmat){ if (p == NULL || invmat == NULL) { return 1 } if (num >10) { return 2 } MatrixOpp(p, num, num, invmat) return 0}int main(){ int i, j int num double *arr=NULL double *result=NULL int flag printf("请输入矩阵维数：\n") scanf("%d",&num) arr=(double *)malloc(sizeof(double)*num*num) result=(double *)malloc(sizeof(double)*num*num) printf("请输入%d*%d矩阵：\n", num, num) for (i = 0i <numi++) { for (j = 0j <numj++) { scanf("%lf", &arr[i * num + j]) } } flag = matrix_inv(arr, num, result) if(flag==0) { printf("逆矩阵为：\n") for (i = 0i <num * numi++) { printf("%lf\t ", *(result + i)) if (i % num == (num - 1)) printf("\n") } } else if(flag==1) { printf("p/q为空\n") } else { printf("超过最大维数\n") } system("PAUSE") free(arr) free(result) return 0}

下面是实现Gauss-Jordan法实矩阵求逆。

#include <stdlib.h>

#include <math.h>

#include <stdio.h>

int brinv(double a[], int n)

{ int *is,*js,i,j,k,l,u,v

double d,p

is=malloc(n*sizeof(int))

js=malloc(n*sizeof(int))

for (k=0k<=n-1k++)

{ d=0.0

for (i=ki<=n-1i++)