求一个四乘四矩阵的逆矩阵

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>

int GJ(int, double **);
double **TwoArrayAlloc(int, int);
void TwoArrayFree(double **);

void main()
{
    int i, j, n;
    double **a;
    n = 4;
    a = TwoArrayAlloc(n, n);

    a[0][0] = 5;    a[0][1] = 7;    a[0][2] = 6;    a[0][3] = 5;
    a[1][0] = 7;    a[1][1] = 10;    a[1][2] = 8;    a[1][3] = 7;
    a[2][0] = 6;    a[2][1] = 8;    a[2][2] = 10;    a[2][3] = 9;
    a[3][0] = 5;    a[3][1] = 7;    a[3][2] = 9;    a[3][3] = 10;
    if (!GJ(n, a))
    {
        printf("矩阵求逆失败
");
        exit(1);
    }
    printf("该矩阵的逆为：
");
    for (i = 0; i<n; i++)
    {
        for (j = 0; j<n; j++)
            printf("%.2f	", a[i][j]);
        printf("
");
    }
}

int GJ(int n, double **a)
{
    int i, j, k;
    double p, q, *h;
    h = (double *)calloc(n, sizeof(double));
    if (h == NULL)
    {
        printf("内存分配失败
");
        exit(1);
    }
    for (k = n; k >= 1; k--)
    {
        p = a[0][0];
        if (p <= 0)
        {
            free(h);
            return (0);
        }
        for (i = 2; i <= n; i++)
        {
            q = a[i - 1][0];
            if (i>k)
                h[i - 1] = q / p;
            else
                h[i - 1] = -q / p;
            for (j = 2; j <= i; j++)
                a[i - 2][j - 2] = a[i - 1][j - 1] + q*h[j - 1];
        }
        a[n - 1][n - 1] = 1 / p;
        for (i = 2; i <= n; i++)
            a[n - 1][i - 2] = h[i - 1];
    }
    free(h);
    return(1);
}

double **TwoArrayAlloc(int r, int c)
{
    double *x, **y;
    int n;
    x = (double *)calloc(r*c, sizeof(double));
    y = (double **)calloc(r, sizeof(double*));
    for (n = 0; n <= r - 1; ++n)
        y[n] = &x[c*n];
    return (y);
}

void TwoArrayFree(double **x)
{
    free(x[0]);
    free(x);
}