计算逆矩阵

/*
* 时间复杂度 O(n^3)
* 输入 a  原矩阵
*     c  逆矩阵
*     n  矩阵的阶数
* 
* 函数说明：将原矩阵a和一个单位矩阵E作成一个大矩阵(a,E)，
* 用初等变换将大矩阵中的a变成E，则会得到(E,a-1)的形式
*/
vector<double> operator * (vector<double> a, double b){
    int n = a.size();
    vector<double>res (n, 0);
    for (int i = 0; i < n; i++){
        res[i] = a[i]*b;
    }
    return res;
}
vector<double> operator - (vector<double>a, vector<double> b){
    int n = a.size();
    vector<double>  res(n, 0);
    for (int i = 0; i < n; i++){
        res[i] = a[i]-b[i];
    }
    return res;
}
void inverse(vector<double> a[], vector<double> c[], int n){
    for (int i = 0; i < n; i++){
        c[i] = vector<double>(n, 0);
    }
    for (int i = 0; i < n; i++){
        c[i][i] = 1;
    }
    for (int i = 0; i < n; i++){
        for (int j = i; j < n; j++){
            if (fabs(a[j][i]) > eps){
                swap(a[i], a[j]);
                swap(c[i], c[j]);
                break;
            }
        }
        c[i] = c[i] * (1.0/a[i][i]);
        a[i] = a[i] * (1.0/a[i][i]);
        for (int j = 0; j < n; j++){
            if (i != j && fabs(a[j][i]) > eps){
                c[j] = c[j] - c[i] * a[j][i];
                a[j] = a[j] - a[i] * a[j][i];
            }
        }
    }
}