下面是實現Gauss-Jordan法實矩陣求逆。

#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=0; k<=n-1; k++)

{ d=0.0;

for (i=k; i<=n-1; i++)

for (j=k; j<=n-1; j++)

{ l=i*n+j; p=fabs(a[l]);

if (p>d) { d=p; is[k]=i; js[k]=j;}

}

if (d+1.0==1.0)

{ free(is); free(js); printf("err**not inv\

");

return(0);

}

if (is[k]!=k)

for (j=0; j<=n-1; j++)

{ u=k*n+j; v=is[k]*n+j;

p=a[u]; a[u]=a[v]; a[v]=p;

}

if (js[k]!=k)

for (i=0; i<=n-1; i++)

{ u=i*n+k; v=i*n+js[k];

p=a[u]; a[u]=a[v]; a[v]=p;

}

l=k*n+k;

a[l]=1.0/a[l];

for (j=0; j<=n-1; j++)

if (j!=k)

{ u=k*n+j; a[u]=a[u]*a[l];}

for (i=0; i<=n-1; i++)

if (i!=k)

for (j=0; j<=n-1; j++)

if (j!=k)

{ u=i*n+j;

a[u]=a[u]-a[i*n+k]*a[k*n+j];

}

for (i=0; i<=n-1; i++)

if (i!=k)

{ u=i*n+k; a[u]=-a[u]*a[l];}

}

for (k=n-1; k>=0; k--)

{ if (js[k]!=k)

for (j=0; j<=n-1; j++)

{ u=k*n+j; v=js[k]*n+j;

p=a[u]; a[u]=a[v]; a[v]=p;

}

if (is[k]!=k)

for (i=0; i<=n-1; i++)

{ u=i*n+k; v=i*n+is[k];

p=a[u]; a[u]=a[v]; a[v]=p;

}

}

free(is); free(js);

return(1);

}

void brmul(double a[], double b[],int m,int n,int k,double c[])

{ int i,j,l,u;

for (i=0; i<=m-1; i++)

for (j=0; j<=k-1; j++)

{ u=i*k+j; c[u]=0.0;

for (l=0; l<=n-1; l++)

c[u]=c[u]+a[i*n+l]*b[l*k+j];

}

return;

}

int main()

{ int i,j;

static double a[6][6]={ {0.2368,0.2471,0.2568,1.2671,0.2568,1.2671},

{1.1161,0.1254,0.1397,0.1490,1.1161,0.1254},

{0.1582,1.1675,0.1768,0.1871,0.1582,1.1675},

{0.1968,0.2071,1.2168,0.2271,0.2071,1.2168},

{0.2368,0.1254,0.1397,0.1490,1.1161,0.2568},

{0.2568,0.2071,1.2168,0.2368,0.2071,1.2168}

};

static double b[6][6],c[6][6];

for (i=0; i<6; i++)

for (j=0; j<6; j++)

b[i][j]=a[i][j];

i=brinv(a,6);

if (i!=0)

{ printf("MAT A IS:\

");

for (i=0; i<6; i++)

{ for (j=0; j<6; j++)

printf("%13.7e ",b[i][j]);

printf("\

");

}

printf("\

");

printf("MAT A- IS:\

");

for (i=0; i<6; i++)

{ for (j=0; j<6; j++)

printf("%13.7e ",a[i][j]);

printf("\

");

}

printf("\

");

printf("MAT AA- IS:\

");

brmul(b,a,6,6,6,c);

for (i=0; i<6; i++)

{ for (j=0; j<6; j++)

printf("%13.7e ",c[i][j]);

printf("\

");

}

}

}