如何找到行列式

Question

double Determinant(double *X, int N){

    /*Solution*/

} 

int main(void)
{

  double X[] = {3, 1, 2, 
                7, 9, 2, 
                4, 6, 9};

  if (Determinant(X,3) == 164) 
   {
    printf("✓");   
   } 

 }

如何求一维数组NxN行列式矩阵？有人能帮我吗？提前致谢。

Answer 1

对于 N > 2，行列式通常以递归形式计算为 SUM(a_i0 * det(X_i * ( -1)ⁱ) 其中X_i是去掉第一列第i行得到的子矩阵，在C语言中，可以写成：

double Determinant(double *X, int N) {
    if (N == 2) {                         // trivial for a 2-2 matrix
        return X[0] * X[3] - X[1] * X[2];
    }
    // allocate a sequential array for the sub-matrix
    double *Y = malloc((N - 1) * (N - 1) * sizeof(double));
    // recursively compute the determinant
    double det = 0.;
    for (int k = 0, s = 1; k < N; k++) {
        // build the submatrix
        for (int i = 0, l=0; i < N; i++) {
            if (i == k) continue;
            for (int j = 1; j < N; j++) {
                Y[l++] = X[j + i * N];
            }
        }
        det += X[k * N] * Determinant(Y, N - 1) * s;
        s = -s;
    }
    free(Y);                        // do not forget to de-alloc...
    return det;
}

如何找到行列式

how to find determinant

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arrays

dimensional

determinants