数学 - Python - 線型代数 - 行列式 - 行列式の計算(連立一次方程式の解法)

線型代数入門

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• Nebo(Windows アプリ)
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• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

線型代数入門(松坂 和夫(著)、岩波書店)の第5章(行列式)、6(行列式の計算)、問題6.を取り組んでみる。

6. 1. 
      
      $\begin{array}{}\det (\begin{array}{ccc}3& 1& 1\\ 4& 3& -1\\ 5& 4& 1\end{array})\end{array}=\det (\begin{array}{ccc}0& 0& 1\\ 7& 4& -1\\ 2& 3& 1\end{array})\\ =21-8\\ =13$
      $\begin{array}{}x\\ =\frac{1}{13}\det (\begin{array}{ccc}-5& 1& 1\\ -2& 3& -1\\ 6& 4& 1\end{array})\end{array}=\frac{1}{13}\det (\begin{array}{ccc}0& 0& 1\\ -7& 4& -1\\ 11& 3& 1\end{array})\\ =\frac{1}{13}\left(-21-44\right)\\ =-\frac{65}{13}\\ =-5$
      $\begin{array}{}y\\ =\frac{1}{13}\det (\begin{array}{ccc}3& -5& 1\\ 4& -2& -1\\ 5& 6& 1\end{array})\end{array}=\frac{1}{13}\det (\begin{array}{ccc}0& 0& 1\\ 7& -7& -1\\ 2& 11& 1\end{array})\\ =\frac{1}{13}\left(77+14\right)\\ =\frac{91}{13}\\ =7$
      $\begin{array}{}z\\ =\frac{1}{13}\det (\begin{array}{ccc}3& 1& -5\\ 4& 3& -2\\ 5& 4& 6\end{array})\end{array}=\frac{1}{13}\det (\begin{array}{ccc}0& 1& 0\\ -5& 3& 13\\ -7& 4& 26\end{array})\\ =\frac{1}{13}\left(130-91\right)\\ =10-7\\ =3$
   2. 
      
      $\begin{array}{}\det (\begin{array}{cccc}1& 2& -1& -1\\ 1& -1& 1& 1\\ 1& -1& -1& 0\\ 1& 0& 0& 1\end{array})\end{array}=\det (\begin{array}{cccc}2& 2& -1& 0\\ 0& -1& 1& 0\\ 1& -1& -1& 0\\ 0& 0& 0& 1\end{array})\\ =\left(2+2\right)-\left(-2+1\right)\\ =5$
      $\begin{array}{}x\\ =\frac{1}{5}\det (\begin{array}{cccc}3& 2& -1& -1\\ 3& -1& 1& 1\\ 4& -1& -1& 0\\ 4& 0& 0& 1\end{array})\end{array}=\frac{1}{5}\det (\begin{array}{cccc}6& 1& 0& 0\\ 3& -1& 1& 1\\ 4& -1& -1& 0\\ 4& 0& 0& 1\end{array})\\ =\frac{1}{5}\det (\begin{array}{cccc}6& 1& 0& 0\\ -1& -1& 1& 0\\ 4& -1& -1& 0\\ 4& 0& 0& 1\end{array})\\ =\frac{1}{5}\left(\left(6+4\right)-\left(-6+1\right)\right)\\ =3$
      $\begin{array}{}y\\ =\frac{1}{5}\det (\begin{array}{cccc}1& 3& -1& -1\\ 1& 3& 1& 1\\ 1& 4& -1& 0\\ 1& 4& 0& 1\end{array})\end{array}=\frac{1}{5}\det (\begin{array}{cccc}2& 7& -1& 0\\ 0& -1& 1& 0\\ 1& 4& -1& 0\\ 1& 4& 0& 1\end{array})\\ =\frac{1}{5}\left(\left(2+7\right)-\left(8+1\right)\right)\\ =0$
      $\begin{array}{}z\\ =\frac{1}{5}\det (\begin{array}{cccc}1& 2& 3& -1\\ 1& -1& 3& 1\\ 1& -1& 4& 0\\ 1& 0& 4& 1\end{array})\end{array}=\frac{1}{5}\det (\begin{array}{cccc}2& 2& 7& 0\\ 0& -1& -1& 0\\ 1& -1& 4& 0\\ 1& 0& 4& 1\end{array})\\ =\frac{1}{5}\left(\left(-8-2\right)-\left(2-7\right)\right)\\ =-1$
      $\begin{array}{}w\\ =\frac{1}{5}\det (\begin{array}{cccc}1& 2& -1& 3\\ 1& -1& 1& 3\\ 1& -1& -1& 4\\ 1& 0& 0& 4\end{array})\end{array}=\frac{1}{5}\det (\begin{array}{cccc}1& 2& 1& -1\\ 1& -1& 0& -1\\ 1& -1& -2& 0\\ 1& 0& 0& 0\end{array})\\ =-\frac{1}{5}\left(\left(1-2\right)-4\right)\\ =1$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, Matrix, solve

x, y, z, w = symbols('x, y, z, w')
exprs = [
    (
        3 * x + y + z + 5,
        4 * x + 3 * y - z + 2,
        5 * x + 4 * y + z - 6
    ),
    (
        x + 2 * y - z - w - 3,
        x - y + z + w - 3,
        x - y - z - 4,
        x + w - 4
    )
]

for i, expr in enumerate(exprs):
    print(f'({chr(ord("a") + i)})')
    for t in [expr, solve(expr)]:
        pprint(t)
        print()
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample6.py
(a)
(3⋅x + y + z + 5, 4⋅x + 3⋅y - z + 2, 5⋅x + 4⋅y + z - 6)

{x: -5, y: 7, z: 3}


(b)
(-w + x + 2⋅y - z - 3, w + x - y + z - 3, x - y - z - 4, w + x - 4)

{w: 1, x: 3, y: 0, z: -1}


$