python numpy linalg solver: 错误答案

Question

我有问题np.linalg.solve求解（我也尝试过使用scipy）但是对于某些线性系统，答案是错误的。在我的程序中生成的系统示例：

Matrix A:

[[7.03894408e+00 1.34629120e+10 2.00000000e+10 1.14564392e+10 1.82002747e+10 1.73205081e+10] [1.34629120e+10 7.03894408e+00 1.82002747e+10 2.00000000e+10 2.23606798e+10 2.07665597e+10] [2.00000000e+10 1.82002747e+10 7.03894408e+00 1.67705098e+10 1.67705098e+10 2.23606798e+10] [1.14564392e+10 2.00000000e+10 1.67705098e+10 7.03894408e+00 1.73205081e+10 1.34629120e+10] [1.82002747e+10 2.23606798e+10 1.67705098e+10 1.73205081e+10 7.03894408e+00 1.95256242e+10] [1.73205081e+10 2.07665597e+10 2.23606798e+10 1.34629120e+10 1.95256242e+10 7.03894408e+00]]

Vector b:

[5.49316406e+42 9.62786249e+22 5.49316406e+42 8.66507624e+23 1.38770661e+25 7.66220239e+24]

Vector x from x = np.linalg.solve(A, b)

[-4.06597524e+32 2.80218361e+32 -2.68178425e+32 2.82035894e+32 1.75304606e+32 3.82470510e+31]

A*x from np.dot(A, x) that should be equal to b

[ 5.49316406e+42 9.28455029e+26 5.49316406e+42 6.18970020e+26 -6.18970020e+26 -1.23794004e+27]

对于某些系统，在向量b和A*x中有更多相等的元素

some verifications:

conditional number: np.linalg.cond(A) = 11.283698804140434

determinant of A: np.linalg.det(A) = -1.146617874355366e+62

norm of A: np.linalg.norm(A) = 99310120330.20604

norm of inv(A): np.linalg.norm(np.linalg.inv(A)) = 1.6365102872452848e-10

一些系统是正常的，解算器给出了正确的答案。你知道吗

谢谢你。你知道吗

Answer 1

你得到的答案不是错的。实际上，比较原始的b和重构的br。。。你知道吗

A
# array([[7.03894408e+00, 1.34629120e+10, 2.00000000e+10, 1.14564392e+10,
#         1.82002747e+10, 1.73205081e+10],
#        [1.34629120e+10, 7.03894408e+00, 1.82002747e+10, 2.00000000e+10,
#         2.23606798e+10, 2.07665597e+10],
#        [2.00000000e+10, 1.82002747e+10, 7.03894408e+00, 1.67705098e+10,
#         1.67705098e+10, 2.23606798e+10],
#        [1.14564392e+10, 2.00000000e+10, 1.67705098e+10, 7.03894408e+00,
#         1.73205081e+10, 1.34629120e+10],
#        [1.82002747e+10, 2.23606798e+10, 1.67705098e+10, 1.73205081e+10,
#         7.03894408e+00, 1.95256242e+10],
#        [1.73205081e+10, 2.07665597e+10, 2.23606798e+10, 1.34629120e+10,
#         1.95256242e+10, 7.03894408e+00]])
b
# array([5.49316406e+42, 9.62786249e+22, 5.49316406e+42, 8.66507624e+23,
#        1.38770661e+25, 7.66220239e+24])

br = A@np.linalg.solve(A,b)

abserr = np.sqrt((br-b)@(br-b))
relerr = abserr / np.sqrt(b@b) 
relerr
# 4.914258035606803e-16

。。。我们得到的相对误差大约是机器精度的2.2倍。。。你知道吗

np.finfo(float).eps
# 2.220446049250313e-16

。。。这其实很准确。你知道吗