Question

我正在尝试求解一个具有6x6矩阵的项（元素）的矩阵

我尝试将gen的逆乘以求解矩阵，但是我不相信答案的正确性。

from sympy import Eq, solve_linear_system, Matrix,count_ops,Mul,horner
import sympy as sp

a, b, c, d, e,f = sp.symbols('a b c d e f')

ad = Matrix(([43.4,26.5,115,-40.5,52.4,0.921],
  [3.78,62.9,127,-67.6,110,4.80],
  [41.25,75.0,213,-88.9, 131, 5.88],
  [-10.6,-68.4,-120,64.6,-132,-8.49],
  [6.5,74.3,121,-72.8,179,29.7],
  [1.2,30.7,49.7,-28.7,91,29.9]))
fb= Matrix(([1,0,0,0,0,0],
        [0,1,0,0,0,0],
        [0,0,1,0,0,0],
        [0,0,0,1,0,0],
        [0,0,0,0,1,0],
        [0,0,0,0,0,1]))

ab = Matrix(([-0.0057],
        [0.0006],
        [-0.0037],
        [0.0009],
        [0.0025],
        [0.0042]))
az = sp.symbols('az')
bz = sp.symbols('bz')
fz = sp.symbols('fz')

gen = Matrix(([az, fz, 0, 0, 0, 0,bz],
            [fz,az,fz,0,0,0,bz],
            [0,fz,az,fz,0,0,bz],
                [0,0,fz,az,fz,0,bz],
                [0,0,0,fz,az,fz,bz],
                [0,0,0,0,fz,az,bz]))

answer = solve_linear_system(gen,a,b,c,d,e,f)
first_solution = answer[a]
df = count_ops(first_solution)
print(df,first_solution)
disolved = zip(first_solution.simplify().as_numer_denom(),(1,-1))
dft = Mul(*[horner(b)**e for b,e in disolved])
dff = count_ops(dft)
print(dff,dft)
_1st_solution = dft.subs({az:ad,fz:fb,bz:ab},simultaneous = True).doit()
print(_1st_solution)

运行代码时，它引发了sympy.matrices.common.ShapeError

Answer 1

在将horner与包含实际上是不可交换的可交换符号的表达式一起使用时（在您的情况下，因为它们代表矩阵），必须小心。您的dft表达式是

(az**2*bz - bz*fz**2)/(az*(az*(az + fz) - 2*fz**2) - fz**3)

但应该是

(az**2 - fz**2)*(az*(az*(az + fz) - 2*fz**2) - fz**3)**(-1)*bz

如果将符号创建为不可交换的，您将收到正确的表达式（如下所示）。

但是您不能将horner与非交换符号一起使用，因此我只是手动重新排列了该表达式；您将必须检查订购是否正确。作为手动进行分解的替代方法，您也可以尝试使用factor_nc来帮助您-但它不能像表达式分解一样处理horner：

>>> ax, bz, fz = symbols('az bz fz, commutative=False)
>>> (az**2*bz - fz**2*bz)
az**2*bz - fz**2*bz
>>> factor_nc(_)
(az**2 - fz**2)*bz

使用sympy求解线性系统6x6矩阵

1 个答案: