computing determinant of a matrix (nxn) recursively

Question

I\'m about to write some code that computes the determinant of a square matrix (nxn), using the Laplace algorithm (Meaning recursive algorithm) as written Wikipedia\'s Lapla

Answer 1

i posted this code because i couldn't fine it on the internet, how to solve n*n determinant using only standard library. the purpose is to share it with those who will find it useful. i started by calculating the submatrix Ai related to a(0,i). and i used recursive determinant to make it short.

  def submatrix(M, c):
    B = [[1] * len(M) for i in range(len(M))]


    for l in range(len(M)):
        for k in range(len(M)):
            B[l][k] = M[l][k]

    B.pop(0)

    for i in range(len(B)):
        B[i].pop(c)
    return B


def det(M):
    X = 0
    if len(M) != len(M[0]):
        print('matrice non carrée')
    else:
        if len(M) <= 2:
            return M[0][0] * M[1][1] - M[0][1] * M[1][0]
        else:
            for i in range(len(M)):
                X = X + ((-1) ** (i)) * M[0][i] * det(submatrix(M, i))
    return X

sorry for not commenting before guys :) if you need any further explanation don't hesitate to ask .