逆矩阵问题：给定2个n*n矩阵A和B，设计一个c语言程序，判定A和B是否互逆

#include "stdio.h"
#include "stdlib.h"
double Surplus(double A[],int m,int n);
double * MatrixInver(double A[],int m,int n);
double * MatrixOpp(double A[],int m,int n) /*矩阵求逆*/
{
int i,j,x,y,k;
double *SP=NULL,*AB=NULL,*B=NULL,X,*C;
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=0;i for(j=0;j {for(k=0;k B[k]=A[k];
{for(x=0;x B[i*n+x]=0;
for(y=0;y 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];
}
}
C=MatrixInver(AB,m,n);

return C;
}

double * MatrixInver(double A[],int m,int n) /*矩阵转置*/
{
int i,j;
double *B=NULL;
B=(double *)malloc(m*n*sizeof(double));

for(i=0;i for(j=0;j B[i*m+j]=A[j*n+i];

return B;
}

double Surplus(double A[],int m,int n) /*求矩阵行列式*/
{

int i,j,k,p,r;
double X,temp=1,temp1=1,s=0,s1=0;

if(n==2)
{for(i=0;i for(j=0;j if((i+j)%2) temp1*=A[i*n+j];
else temp*=A[i*n+j];
X=temp-temp1;}
else{
for(k=0;k {for(i=0,j=k;i temp*=A[i*n+j];
if(m-i)
{for(p=m-i,r=m-1;p>0;p--,r--)
temp*=A[r*n+p-1];}
s+=temp;
temp=1;
}

for(k=n-1;k>=0;k--)
{for(i=0,j=k;i=0;i++,j--)
temp1*=A[i*n+j];
if(m-i)
{for(p=m-1,r=i;r temp1*=A[r*n+p];}
s1+=temp1;
temp1=1;
}

X=s-s1;}
return X;
}