求解一個(gè)$m\times n$階實(shí)矩陣$A$的秩。從數(shù)學(xué)角度來說矩陣的秩就是矩陣中線性無關(guān)的行向量的最大個(gè)數(shù)，并且初等行變換和列變換均不會影響矩陣的秩的大小，為了實(shí)現(xiàn)這一過程常使用全選的高斯消去法將矩陣變?yōu)樯先蔷仃?#xff0c;則其中非零的行向量個(gè)數(shù)就是矩陣的秩。

#include "stdio.h" #include "stdlib.h" #include "math.h"double mat[100], eps; int m,n; int r_mrank(double *mat,int m, int n,double eps){int i,j,k,p,is,js,l,v,rank;double tmp, pivot;double *cpmat;if(mat == NULL){printf("matrix pointer is Null.\n");return(0);}cpmat = (double*)malloc(m*n*sizeof(double)); for(i=0; i<m*n; i++)cpmat[i] = mat[i];p = m<n?m:n;rank = 0;for(k=0; k<p; k++){pivot = 0.0;for(i=k; i<m; i++)for(j=k; j<n; j++){tmp = fabs(cpmat[i*n+j]);if(tmp > pivot){pivot = tmp;is = i;js = j;}}if(pivot < eps) return(rank);rank++; if(is != k)for(j=k; j<n; j++){l = k*n + j;v = is*n + j;tmp = cpmat[l];cpmat[l] = cpmat[v];cpmat[v] = tmp;}if(js != k)for(i=k; i<m; i++){l = i*n + k;v = i*n + js;tmp = cpmat[l];cpmat[l] = cpmat[v];cpmat[v] = tmp;}for(i=k+1; i<m; i++)tmp = cpmat[i*n+k]/cpmat[k*n+k];for(j=k+1; j<n; j++)cpmat[i*n+j] -= tmp*cpmat[k*n+j];}}free(cpmat);return(rank); }int main(){int count, rank, i ,j;scanf("%d %d %lf",&m,&n,&eps);printf("Input the matric please: \n");for(count=0;count<m*n;count++){scanf("%lf", mat+count);}printf("The matrix you input is \n");for(i=0;i<m;i++){for(j=0;j<n;j++)printf("%lf ", *(mat+i*n + j));printf("\n");}rank = r_mrank(mat,m,n,eps);printf("the rank of the matrix is: %d", rank);return 0; }

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