連立1次方程式
次のような連立１次方程式を考える．
\begin{align}
&a_{11} x_1 + a_{12} x_2 + a_{13} x_3 + \cdots \ \ + a_{1n} x_n = b_1
\nonumber \\
&a_{21} x_1 + a_{22} x_2 + a_{23} x_3 + \cdots \ \ + a_{2n} x_n = b_2
\nonumber \\
&a_{31} x_1 + a_{32} x_2 + a_{33} x_3 + \cdots \ \ + a_{3n} x_n = b_3
\nonumber \\
&\cdots \nonumber \\
&a_{n1} x_1 + a_{n2} x_2 + a_{n3} x_3 + \cdots \ \ + a_{nn} x_n = b_n
\nonumber
\end{align}
これを行列表示すると，
$$
\begin{pmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\
a_{31} & a_{32} & a_{33} & \cdots & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n
\end{pmatrix}
=
\begin{pmatrix}
b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n
\end{pmatrix}
$$
あるいは，
$$
[a] (x) = (b)
$$
ここで，
$$
[a] =
\begin{pmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\
a_{31} & a_{32} & a_{33} & \cdots & a_{3n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & a_{n3} & \cdots & a_{nn}
\end{pmatrix}, \ \ \ \ \
(x) =
\begin{pmatrix}
x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n
\end{pmatrix}, \ \ \ \ \
(b) =
\begin{pmatrix}
b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n
\end{pmatrix}
$$
モジュールのインポート
import numpy as np
from scipy import linalg
Numpy
https://numpy.org/doc/stable/reference/generated/numpy.linalg.solve.html
a = np.array([[3.0, 2.0, 1.0],[5.0, -5.0, 4.0],[6.0, 0.0, 1.0]])
a
array([[ 3., 2., 1.],
[ 5., -5., 4.],
[ 6., 0., 1.]])
x = np.array([1.0, 2.0, 3.0])
b = a@x
b
array([10., 7., 9.])
x = np.linalg.solve(a,b) # numpy
x
array([1., 2., 3.])
a@x
array([10., 7., 9.])
np.allclose(np.dot(a, x), b) # 解の確認
True
np.dot(a, x) - b
array([-1.77635684e-15, 0.00000000e+00, 0.00000000e+00])
np.dot(a, x) == b
array([False, True, True])
Scipy
https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.solve.html
a2 = np.array([[3,2,3,10],[2,-2,5,8],[3,3,4,9],[3,4,-3,-7]])
a2
array([[ 3, 2, 3, 10],
[ 2, -2, 5, 8],
[ 3, 3, 4, 9],
[ 3, 4, -3, -7]])
x2 = np.array([4,3,2,1])
x2
array([4, 3, 2, 1])
b2 = np.array([4,1,3,2])
b2 = a2@x2
b2
array([34, 20, 38, 11])
x2 = linalg.solve(a2,b2)# scipy
x2
array([4., 3., 2., 1.])
np.allclose(np.dot(a2, x2), b2) # 解の確認
True
np.dot(a2, x2) - b2
array([0.00000000e+00, 3.55271368e-15, 0.00000000e+00, 0.00000000e+00])
np.dot(a2, x2) == b2
array([ True, False, True, True])
逆行列
https://numpy.org/doc/stable/reference/generated/numpy.linalg.inv.html
a
array([[ 3., 2., 1.],
[ 5., -5., 4.],
[ 6., 0., 1.]])
ai = np.linalg.inv(a) # 逆行列の計算
ai
array([[-0.09433962, -0.03773585, 0.24528302],
[ 0.35849057, -0.05660377, -0.13207547],
[ 0.56603774, 0.22641509, -0.47169811]])
np.linalg.inv(ai)
array([[ 3.00000000e+00, 2.00000000e+00, 1.00000000e+00],
[ 5.00000000e+00, -5.00000000e+00, 4.00000000e+00],
[ 6.00000000e+00, 6.54612632e-18, 1.00000000e+00]])
ai@b
array([1., 2., 3.])
np.dot(ai,b)
array([1., 2., 3.])
行列式
https://numpy.org/doc/stable/reference/generated/numpy.linalg.det.html
det_a = np.linalg.det(a)
det_a
52.999999999999986
行列のランク(階数）
https://numpy.org/doc/stable/reference/generated/numpy.linalg.matrix_rank.html
np.linalg.matrix_rank(a)
3