練習問題メモ 5（連立1次方程式）

練習問題メモ 5（連立1次方程式）#

5.1#

掃き出し法により、次の1～3の連立1次方程式を解け。また、係数行列および拡大係数行列について、それぞれの階数を求めよ

1.#

\begin{array}{r} (\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array}) (\begin{array}{l} x_{1} \\ x_{2} \end{array}) = (\begin{array}{l} 1 \\ 0 \\ 0 \end{array}) \end{array}

import numpy as np
A = np.array([
    [1, 2, 1],
    [3, 4, 0],
    [5, 6, 0],
], dtype=np.float32)

# 1行目を-2倍して2行目に加える
A[1, :] += A[0, :] * (-2)
A

array([[ 1.,  2.,  1.],
       [ 1.,  0., -2.],
       [ 5.,  6.,  0.]], dtype=float32)

# 1行目を-3倍して3行目に加える
A[2, :] += A[0, :] * (-3)
A

array([[ 1.,  2.,  1.],
       [ 1.,  0., -2.],
       [ 2.,  0., -3.]], dtype=float32)

# 3行目を(1/2)倍する
A[2, :] = (1 / 2) * A[2, :]
A

array([[ 1. ,  2. ,  1. ],
       [ 1. ,  0. , -2. ],
       [ 1. ,  0. , -1.5]], dtype=float32)

# 2行目を-1倍して3行目に加える
A[2, :] += A[1, :] * (-1)
A

array([[ 1. ,  2. ,  1. ],
       [ 1. ,  0. , -2. ],
       [ 0. ,  0. ,  0.5]], dtype=float32)

# 2行目を-1倍して1行目に加える
A[0, :] += A[1, :] * (-1)
A

array([[ 0. ,  2. ,  3. ],
       [ 1. ,  0. , -2. ],
       [ 0. ,  0. ,  0.5]], dtype=float32)

# 1行目を(1/2)倍する
A[0, :] = (1 / 2) * A[0, :]
A

array([[ 0. ,  1. ,  1.5],
       [ 1. ,  0. , -2. ],
       [ 0. ,  0. ,  0.5]], dtype=float32)

a1 = A[0, :].copy()
a2 = A[1, :].copy()
A[0, :] = a2
A[1, :] = a1
A

array([[ 1. ,  0. , -2. ],
       [ 0. ,  1. ,  1.5],
       [ 0. ,  0. ,  0.5]], dtype=float32)

\begin{array}{r} (\begin{array}{ccc} 1 & 0 & - 2 \\ 0 & 1 & 3 / 2 \\ 0 & 0 & 1 / 2 \end{array}) \end{array}

3行目も零ベクトルにできなかった → 不能？

1-2. 係数行列の階数

\begin{array}{r} (\begin{array}{c} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}) \end{array}

と、階段行列の形であり零の行ベクトルでない行数は2なので2

1-3. 拡大係数行列の階数

（3行目を2倍すれば対角成分が1になるので）

3

2.#

$\begin{array}{r} (\begin{array}{ccc} 1 & 1 & - 2 \\ 1 & - 2 & 1 \\ - 2 & 1 & 1 \end{array}) (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}) = (\begin{array}{c} 0 \\ 3 \\ - 3 \end{array}) \end{array}$

import numpy as np
A = np.array([
    [1, 1, -2, 0],
    [1, -2, 1, 3],
    [-2, 1, 1, -3],
])

# 1行目を2行目から引く
A[1, :] += A[0, :] * (-1)
A

array([[ 1,  1, -2,  0],
       [ 0, -3,  3,  3],
       [-2,  1,  1, -3]])

# 1行目を2倍して3行目に加える
A[2, :] += A[0, :] * (2)
A

array([[ 1,  1, -2,  0],
       [ 0, -3,  3,  3],
       [ 0,  3, -3, -3]])

# 2行目を3行目に加える
A[2, :] += A[1, :] * (1)
A

array([[ 1,  1, -2,  0],
       [ 0, -3,  3,  3],
       [ 0,  0,  0,  0]])

# 2行目を1/3する
A[1, :] = A[1, :] * (-1/3)
A

array([[ 1,  1, -2,  0],
       [ 0,  1, -1, -1],
       [ 0,  0,  0,  0]])

# 2行目を1行目から引く
A[0, :] += A[1, :] * (-1)
A

array([[ 1,  0, -1,  1],
       [ 0,  1, -1, -1],
       [ 0,  0,  0,  0]])

\begin{array}{r} {\begin{cases} x_{1} - x_{3} = 1 \\ x_{2} - x_{3} = - 1 \end{cases} \end{array}

$x_{3}$ を任意定数に置き換える

\begin{array}{r} {\begin{cases} x_{1} - c = 1 \\ x_{2} - c = - 1 \end{cases} \end{array}

\begin{array}{r} x_{3} = c かつ {\begin{cases} x_{1} = c + 1 \\ x_{2} = c - 1 \end{cases} \end{array}

3.#

\begin{array}{r} (\begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 0 & 1 & 2 & 7 \end{array}) (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = (\begin{array}{l} 1 \\ 0 \\ 1 \end{array}) \end{array}

from sympy import Matrix
A = Matrix([
    [1, 2, 3, 4, 1],
    [2, 3, 4, 1, 0],
    [0, 1, 2, 7, 1],
])

A[1, :] += A[0, :] * (-2)
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 1 \\ 0 & - 1 & - 2 & - 7 & - 2 \\ 0 & 1 & 2 & 7 & 1 \end{array}] \end{array}

A[2, :] += A[1, :]
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 1 \\ 0 & - 1 & - 2 & - 7 & - 2 \\ 0 & 0 & 0 & 0 & - 1 \end{array}] \end{array}

A[1, :] = A[1, :] * (-1)
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 1 \\ 0 & 1 & 2 & 7 & 2 \\ 0 & 0 & 0 & 0 & - 1 \end{array}] \end{array}

A[0, :] += A[1, :] * (-2)
A

\begin{array}{r} [\begin{array}{c} 1 & 0 & - 1 & - 10 & - 3 \\ 0 & 1 & 2 & 7 & 2 \\ 0 & 0 & 0 & 0 & - 1 \end{array}] \end{array}

3行目を零ベクトルにできなかった → 不能？

係数行列のランク: 2

拡大係数行列のランク: 3

5.2#

掃き出し法により、次の同次連立 1 次方程式を解け。 $ $(\begin{array}{llll} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 0 & 1 & 2 & 7 \end{array}) (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = (\begin{array}{l} 0 \\ 0 \\ 0 \end{array})$ $

from sympy import Matrix
A = Matrix([
    [1, 2, 3, 4, 0],
    [2, 3, 4, 1, 0],
    [0, 1, 2, 7, 0],
])

A[1, :] += A[0, :] * (-2)
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 0 \\ 0 & - 1 & - 2 & - 7 & 0 \\ 0 & 1 & 2 & 7 & 0 \end{array}] \end{array}

A[2, :] += A[1, :]
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 0 \\ 0 & - 1 & - 2 & - 7 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}] \end{array}

A[1, :] = A[1, :] * (-1)
A

\begin{array}{r} [\begin{array}{c} 1 & 2 & 3 & 4 & 0 \\ 0 & 1 & 2 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}] \end{array}

A[0, :] += A[1, :] * (-2)
A

\begin{array}{r} [\begin{array}{c} 1 & 0 & - 1 & - 10 & 0 \\ 0 & 1 & 2 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}] \end{array}

\begin{array}{r} {\begin{cases} x_{1} - x_{3} - 10 x_{4} = 0 \\ x_{2} - 2 x_{3} + 7 x_{4} = 0 \end{cases} \end{array}

$x_{3}, x_{4}$ を任意定数に置き換える

\begin{array}{r} {\begin{cases} x_{1} - c_{1} - 10 c_{2} = 0 \\ x_{2} - 2 c_{1} + 7 c_{2} = 0 \end{cases} \end{array}

\begin{array}{r} {\begin{cases} x_{3} = c_{1} \\ x_{4} = c_{2} \end{cases} かつ {\begin{cases} x_{1} = c_{1} + 10 c_{2} \\ x_{2} = 2 c_{1} - 7 c_{2} \end{cases} \end{array}

5.3#

p,q,rを定数とする。連立1次方程式 $ $(\begin{array}{ccc} - 2 & 1 & 1 \\ 1 & - 2 & 1 \\ 1 & 1 & - 2 \end{array}) (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}) = (\begin{array}{l} p \\ q \\ r \end{array})$ $ の解が存在するためのp,q,rの条件を求めよ。

from sympy import Matrix, symbols
p, q, r = symbols("p q r")
A = Matrix([
    [-2, 1, 1, p],
    [1, -2, 1, q],
    [1, 1, -2, r],
])

# 1行目と2行目を入れ替える
a1 = A[0, :].copy()
a2 = A[1, :].copy()
A[0, :] = a2
A[1, :] = a1
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & 1 & q \\ - 2 & 1 & 1 & p \\ 1 & 1 & - 2 & r \end{array}] \end{array}

A[1,:] += 2 * A[0,:]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & 1 & q \\ 0 & - 3 & 3 & p + 2 q \\ 1 & 1 & - 2 & r \end{array}] \end{array}

A[2,:] += -1 * A[0,:]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & 1 & q \\ 0 & - 3 & 3 & p + 2 q \\ 0 & 3 & - 3 & - q + r \end{array}] \end{array}

A[2,:] += 1 * A[1,:]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & 1 & q \\ 0 & - 3 & 3 & p + 2 q \\ 0 & 0 & 0 & p + q + r \end{array}] \end{array}

$p + q + r \neq 0$ だと両者のRankが違うので解なし

連立一次方程式の定理より、両者のランクが合えば解が存在するので $p + q + r = 0$ なら成り立つ

\begin{array}{r} x_{1} - 2 x_{2} + x_{3} = q \\ - 3 x_{2} + 3 x_{3} = p + 2 q \end{array}

\begin{array}{r} x_{1} - 2 x_{2} + x_{3} = q \\ - x_{2} + x_{3} = (p + 2 q) / 3 \end{array}

検算用

掃き出し法計算機 - Wolfram|Alpha

練習問題メモ 5（連立1次方程式）

Contents

練習問題メモ 5（連立1次方程式）#

5.1#

1.#

2.#

3.#

5.2#

5.3#