c-如何求行列式

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的行列式矩阵？有人能帮我吗？提前谢谢。

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是通过移除第一列和第i行获得的子矩阵。在C语言中，它可以写成：

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;
}