生成100x100矩阵并计算该行列式

Question

在下面的代码中，我为通用矩阵A实现了LU分解。基于此实现，我编写了一个计算矩阵行列式的函数。我已经能够使用代码

为2x2矩阵确定它

mat = [[1,0], [0,1]]

但是，现在我想通过生成100x100矩阵并计算该行列式来对其进行测试。那么如何将其放入代码中呢？我是否需要使用类似的

np.round(np.random.rand(100, 100)*10)

正在为此寻求帮助。这是我下面的代码！

def matrixMul(A, B):
    TB = list(zip(*B))
    return [[sum(ea*eb for ea,eb in zip(a,b)) for b in TB] for a in A]
def pivotize(m):
    """Creates the pivoting matrix for m."""
    n = len(m)
    ID = [[float(i == j) for i in range(n)] for j in range(n)]
    r = 0
    for j in range(n):
        row = max(range(j, n), key=lambda i: abs(m[i][j]))
        if j != row:
            ID[j], ID[row] = ID[row], ID[j]
            r += 1
    return ID, r
def lu(A):
    """Decomposes a nxn matrix A by PA=LU and returns L, U and P."""
    n = len(A)
    L = [[0.0] * n for i in range(n)]
    U = [[0.0] * n for i in range(n)]
    P, r = pivotize(A)
    A2 = matrixMul(P, A)
    for j in range(n):
        L[j][j] = 1.0
        for i in range(j+1):
            s1 = sum(U[k][j] * L[i][k] for k in range(i))
            U[i][j] = A2[i][j] - s1
        for i in range(j, n):
            s2 = sum(U[k][j] * L[i][k] for k in range(j))
           L[i][j] = (A2[i][j] - s2) / U[j][j]
    return (L, U, P, r)
def trace(m):
    n = len(m)
    r = 1
    for i in range(n):
        if len(m[i]) <= i:
            break
        r *= m[i][i]
    return r
def det(m):
    l, u, p, r = lu(m)
    return (-1)**r * trace(l) * trace(u)
mat = [[1,0], [0,1]]
print(det(mat))