我在np.linalg.solve上遇到了问题(我也尝试过使用scipy(,但对于某些线性系统来说,答案是错误的。在我的程序中生成的系统示例:
矩阵 A:
[[7.03894408e+00 1.34629120e+10 2.00000000e+101.14564392e+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]]
向量 b:
[5.49316406e+42 9.62786249e+225.49316406e+42 8.66507624e+23 1.38770661e+25 7.66220239e+24]
向量 x 从 x = np.linalg.solve(A, b(
[-4.06597524e+32 2.80218361e+32 -2.68178425e+32 2.82035894e+321.75304606e+32 3.82470510e+31]
来自 np.dot(A, x( 的 A*x,应等于 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,向量 b 和 A*x 中相等的元素数量更多
一些验证:
条件编号:np.linalg.cond(A( = 11.283698804140434
A的行列式:np.linalg.det(A( = -1.146617874355366e+62
A的范数: np.linalg.norm(A( = 99310120330.20604
inv(A( 的范数: np.linalg.norm(np.linalg.inv(A(( = 1.6365102872452848e-10
有些系统是可以的,求解器给出正确的答案。
谢谢。
你得到的答案没有错。事实上,将原始b
与重建的br
进行比较......
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约为机器精度的 2.2 倍......
np.finfo(float).eps
# 2.220446049250313e-16
。这实际上是相当准确的。