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== 連立方程式:掃き出し法,行基本変形 ==

○ この頁では,主に連立方程式の「掃き出し法」による解き方(または「行基本変形」による解き方,または「ガウスの消去法」による解き方)を扱います.

○ 全体の流れ
【未知数2個,方程式2個のとき】
 次の連立方程式において,未知数xを消去したいとき,(1)式のように≪xの係数が1≫になっていると,2倍すれば(2)式のxの係数と簡単にそろえることができ,引き算によってxを消去できます.

1x+3y=5…(1)
2x+5y=9…(2)
(2)−(1)×2
1x+3y=5…(1)
−y=−1…(3)
同様にして,(3)式の≪yの係数を1≫にしておくと,(1)式からyを消去できます.そのためには,(3)式をyの係数−1で割るとよい.

x+3y=5…(1)
1y=1…(3’)
(1)−(3’)×3
x=2…(1’) …(←これが答)
y=1…(3’)


【未知数3個,方程式3個のとき】
なお,各式は変形するたびに書き変わりますが,ここでは記号を繁雑にしないために,「書き換えた結果」も新たに(1)(2)(3)で表すものとします.
2x+y−z=1…(1)
x−y+2z=5…(2)
−2x+3y+z=7…(3)
(1)の≪xの係数を1≫にするために,(1)をその係数2で割る方法もあるが,ここでは「行基本変形」のもう一つの柱を解説するために,他の方法を示す.すなわち,(2)のようにすでに≪xの係数が1≫になっているものがあれば,単に(1)と(2)を入れ替えるだけで(1)のxの係数が1になる.このように,連立方程式を解くときに,必要に応じて「行を入れ替えてもよい」.
1x−y+2z=5…(1)
2x+y−z=1…(2)
−2x+3y+z=7…(3)
(2)−(1)×2
(3)−(1)×(−2)により(2)(3)のxを消去する
1x−y+2z=5…(1)
3y−5z=−9…(2)
y+5z=17…(3)
(2)の≪yの係数を1≫にするために,(2)をその係数3で割る方法もあるが,ここでは(3)の≪yの係数が1≫になっているので,単に(2)と(3)を入れ替える.
1x−y+2z=5…(1)
1y+5z=17…(2)
3y−5z=−9…(3)
(1)−(2)×(−1)
(3)−(2)×3により(1)(3)のyを消去する
1x+7z=22…(1)
1y+5z=17…(2)
−20z=−60…(3)
(3)の≪zの係数を1≫にするために,(3)をその係数−20で割る
1x+7z=22…(1)
1y+5z=17…(2)
1z=3…(3)
(1)−(3)×7
(2)−(3)×5により(1)(2)のzを消去する
1x=1…(1)
1y=2…(2)
1z=3…(3) (←これが答)

←【なぜ係数を1にするのか】
 次のような未知数2個の連立方程式なら,中学校で習ったように,(1)×3−(2)×2xを消去できるのではないか?と思う人もあるかもしれない.
2x+3y=1…(1)
3x+5y=2…(2)
 確かに,未知数2個ぐらいなら「目で見ながら」 (1)×n−(2)×m のような係数をていねいに考えていくことができるが,次のように未知数の多い連立方程式では「自分の係数で割る=係数を1にする」と決めると,「常に機械的に処理」できるようになります.
なお,各式の係数は変形するたびに書き変わりますが,ここでは記号を繁雑にしないために,「書き換えた結果」も新たにamnなどで表すものとします.
a11x1+a12x2+a13x3+a14x4+...=p1…(1)
a21x1+a22x2+a23x3+a24x4+...=p2…(2)
a31x1+a32x2+a33x3+a34x4+...=p3…(3)
a41x1+a42x2+a43x3+a44x4+...=p4…(4)
……(n)

1x1+a12x2+a13x3+a14x4+...=p1…(1)
a21x1+a22x2+a23x3+a24x4+...=p2…(2)
a31x1+a32x2+a33x3+a34x4+...=p3…(3)
a41x1+a42x2+a43x3+a44x4+...=p4…(4)
……(n)
(2)−a21×(1)
(3)−a31×(1)
(4)−a41×(1) によって(1)以外のx1を消します

1x1+a12x2+a13x3+a14x4+...=p1…(1)
a22x2+a23x3+a24x4+...=p2…(2)
a32x2+a33x3+a34x4+...=p3…(3)
a42x2+a43x3+a44x4+...=p4…(4)
……(n)
これで(1)以外のx1の係数が0になったら 次は(2)をx2の係数で割って,その係数を1にします.
1x1+a12x2+a13x3+a14x4+...=p1…(1)
1x2+a23x3+a24x4+...=p2…(2)
a32x2+a33x3+a34x4+...=p3…(3)
a42x2+a43x3+a44x4+...=p4…(4)
……(n)
(1)−a12×(2)
(3)−a32×(2)
(4)−a42×(2) によって(2)以外のx2を消します

1x1+a13x3+a14x4+...=p1…(1)
1x2+a23x3+a24x4+...=p2…(2)
+a33x3+a34x4+...=p3…(3)
+a43x3+a44x4+...=p4…(4)
……(n)
次は(3)式を使ってx3を,さらに(4)式を使ってx4を,...と処理していくと,次々に解けていくことになります.

【行列の形で書くとき】
 はじめの未知数2個の連立方程式を行列の形で書くと,次のようになります.
1
2
3
5
x
y
= 5
9
最後の解を表す形を,行列の形で書くと,次のようになります.
1
0
0
1
x
y
= 2
1
 未知数3個の連立方程式は
1
2
−2
−1
1
3
2
−1
1
x
y
z
= 5
1
7
最後の解を表す形を,行列の形で書くと,次のようになります.
1
0
0
0
1
0
0
0
1
x
y
z
= 1
2
3
 このように,行列の形で考えると,各々の行を「定数倍したり」「足したり引いたり」「行を入れ替えたりして」,左辺の係数行列を単位行列にすれば,右辺の係数行列が解を表すことになります.
【2×2の連立方程式の解き方:要約】
■問題の形
a
b
c
d
x
y
= p
q

■この形にすれば答
1
0
0
1
x
y
= r
s
左辺の係数行列に右辺の定数項を付けたもの:「拡大係数行列」で表すと
■問題の形
a
b
c
d
p
q

■この形にすれば答
1
0
0
1
r
s
※ 逆行列や階数の項目には「列基本変形」という用語もありますが,ここで扱っている連立方程式の解き方については「行基本変形」だけで行うことが重要.ここまでの流れを見て分かるように,もし「列基本変形」をやってしまうと,x, yの解が入れ替わってしまうので,元の方程式の解ではなくなります.
【3×3の連立方程式の解き方:要約】
■問題の形
a
d
g
b
e
h
c
f
i
x
y
z
= p
q
r

■この形にすれば答
1
0
0
0
1
0
0
0
1
x
y
z
= u
v
w
左辺の係数行列に右辺の定数項を付けたもの:「拡大係数行列」で表すと
■問題の形
a
d
g
b
e
h
c
f
i
p
q
r

■この形にすれば答
1
0
0
0
1
0
0
0
1
u
v
w

【具体例による解説】 …3×3の場合について
[1]
A) 1行1列成分が0でなければ,1行目をその成分で割る
図1のように,他の成分は分数になっても構わずに割ります
図1
2
3
4
1
4
−1
−1
−5
1
1
−4
5
1
3
4
1/2
4
−1
−1/2
−5
1
1/2
−4
5

B) 1行1列成分が0であれば,1列目のうちで0でないものを探し,行を入れ替えてから A)の操作を行う
図2は,1行目と2行目を入れ替えた場合を示しています
図2
0
3
4
1
4
−1
−1
−5
1
−1
−4
5
3
0
4
4
1
−1
−5
−1
1
−4
−1
5

C) 1列目の成分が全部0のときは,xは任意の値になる
図3のような場合,xの係数が全部0だからxは何であっても構いません
図3
0
0
0
1
4
−1
−1
−5
1
3
13
−3


[2]
以下において,連立方程式のときと同様に1行目を(1),2行目を(2)などで表す
1
d
g
b
e
h
c
f
i
p
q
r
の形になったら,
(2)−(1)×d
(3)−(1)×g
により係数dg0にします.
図1の右側から出発したとき,図4のようになります
図4
1
3
4
1/2
4
−1
−1/2
−5
1
1/2
−4
5
1
0
0
1/2
5/2
−3
−1/2
−7/2
3
1/2
−11/2
3

以上により,xの係数が処理できたので,次はyの係数(第2列)について,上記の[1][2]の操作を行います

[1]
A) 2行2列成分が0でなければ,2行目をその成分で割る
図4の右側から出発したとき,図5のようになります
図5
1
0
0
1/2
1
−3
−1/2
−7/2
3
1/2
−11/2
3
1
0
0
1/2
1
−3
−1/2
−7/5
3
1/2
−11/5
3

図6
1
0
0
1/2
1
−3
−1/2
−7/5
3
1/2
−11/5
3
1
0
0
0
1
0
1/5
−7/5
−6/5
8/5
−11/5
−18/5

もし,2行2列成分が0でなければ,以下の処理はxのときのB)C)と同様です

[2]
1
0
0
b
1
h
c
f
i
p
q
r
の形になったら,
(1)−(2)×b
(3)−(2)×h
により係数bh0にします.
図5の右側から出発したとき,図6のようになります
以上により,yの係数が処理できたので,次はzの係数(第3列)についても同様に行います
図6の右側から出発したとき,図7のようになり,これは元の連立方程式の解がx=1, y=2, z=3であることを表しています.
図7
1
0
0
0
1
0
1/5
−7/5
−6/5
8/5
−11/5
−18/5
1
0
0
0
1
0
0
0
1
1
2
3


 以下,正しい番号を選択してください.
[問題1]
次の各行列は,拡大係数行列で表された連立方程式を掃き出し法によって解く過程を示したものです.(ア)(イ)(ウ)に入るものを各々右の選択肢から選んでください.

3
2
4
2
1
3
1
3
−2
4
3
5
1
2
4
2/3
1
3
1/3
3
−2
4/3
3
5

1
0
0
2/3
(ア)
1/3
1/3
7/3
−10/3
4/3
1/3
−1/3
1
0
0
2/3
1
1/3
1/3
−7
−10/3
4/3
−1
−1/3

1
0
0
0
1
0
5
−7
(イ)
2
−1
0
1
0
0
0
1
0
5
−7
1
2
−1
0

1
0
0
0
1
0
0
0
1
(ウ)
−1
0



(ア) 1−2/3 2−1/3 31/3 42/3



(イ) 1−7 2−5 3−3 4−1



(ウ) 11 22 33 44



[問題2]
次の各行列は,拡大係数行列で表された連立方程式を掃き出し法によって解く過程を示したものです.(エ)(オ)(カ)に入るものを各々右の選択肢から選んでください.

0
3
−2
1
2
0
3
−4
5
−1
4
−4
3
0
−2
2
1
0
−4
3
5
4
−1
−4

1
0
−2
2/3
1
0
−4/3
3
5
(エ)
−1
−4
1
0
0
2/3
1
4/3
−4/3
3
7/3
4/3
−1
−4/3

1
0
0
0
1
0
−2
3
(オ)
2
−1
0
1
0
0
0
1
0
−2
3
1
2
−1
0

1
0
0
0
1
0
0
0
1
(カ)
−1
0



(エ) 1−4/3 2−2/3 32/3 44/3



(オ) 1−5/3 2−2/3 31/3 44/3



(カ) 1−1 21 32 43


[問題3]
次の各行列は,拡大係数行列で表された連立方程式を掃き出し法によって解く過程を示したものです.(キ)(ク)(ケ)に入るものを各々右の選択肢から選んでください.

1
2
6
−1
−1
3
4
5
−1
1
1
−1
1
0
0
−1
1
9
4
−3
−25
1
−1
(キ)

1
0
0
0
1
0
−1
−3
(ク)
0
−1
2
1
0
0
0
1
0
−1
−3
1
0
−1
1

1
0
0
0
1
0
0
0
1
−1
(ケ)
1


(キ) 1−7 2−5 3−3 4−1



(ク) 10 21 32 43



(ケ) 10 21 32 43



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