Auf solche Gleichungssysteme sind wir beim Untersuchen der anderen Gleichungssysteme schon mehrmals gestoßen; immer dann, wenn der Gauß-Algorithmus so viele Nullzeilen erzeugt, dass letztendlich weniger Zeilen für die Lösung relevant waren als Unbekannte zu finden waren.
Dann kann man eigentlich nicht von einem Gleichungssystem sprechen. Bei zwei Unbekannten sieht eine solche Gleichung wie folgt aus:
a
⋅
x
1
+
b
⋅
x
2
=
e
{\displaystyle a\cdot x_{1}+b\cdot x_{2}=e}
Dann ist
x
1
=
−
b
/
a
⋅
x
2
+
e
/
a
{\displaystyle x_{1}=-b/a\cdot x_{2}+e/a}
x
2
=
t
{\displaystyle x_{2}=t}
x
1
=
−
b
/
a
⋅
t
+
e
/
a
{\displaystyle x_{1}=-b/a\cdot t+e/a}
L
=
{
(
−
b
/
a
⋅
t
+
e
/
a
|
t
)
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,-b/a\cdot t+e/a\,|\,t\,\right),\;t\in \mathbb {R} \right\}}
oder etwas anders geschrieben
L
=
{
(
e
/
a
|
0
)
+
t
⋅
(
−
b
/
a
|
1
)
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{(\,e/a\,|\,0\,)+t\cdot (\,-b/a\,|\,1\,),\;t\in \mathbb {R} \right\}}
Alternativ kann man auch
x
2
{\displaystyle x_{2}}
p
⋅
e
/
b
{\displaystyle p\cdot e/b}
x
1
=
(
1
−
p
)
⋅
e
/
a
{\displaystyle x_{1}=(1-p)\cdot e/a}
L
=
{
(
(
1
−
p
)
⋅
e
/
a
|
p
⋅
e
/
b
)
,
p
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,(1-p)\cdot e/a\,|\,p\cdot e/b\,\right),\;p\in \mathbb {R} \right\}}
a
⋅
x
1
+
b
⋅
x
2
+
c
⋅
x
3
=
e
{\displaystyle a\cdot x_{1}+b\cdot x_{2}+c\cdot x_{3}=e}
Dann ist
x
1
=
−
b
/
a
⋅
x
2
−
c
/
a
⋅
x
3
+
e
/
a
{\displaystyle x_{1}=-b/a\cdot x_{2}-c/a\cdot x_{3}+e/a}
x
2
=
s
{\displaystyle x_{2}=s}
x
3
=
t
{\displaystyle x_{3}=t}
x
1
=
−
b
/
a
⋅
t
−
c
/
a
⋅
s
+
e
/
a
{\displaystyle x_{1}=-b/a\cdot t-c/a\cdot s+e/a}
L
=
{
(
−
b
/
a
⋅
t
−
c
/
a
⋅
s
+
e
/
a
|
s
|
t
)
,
s
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,-b/a\cdot t-c/a\cdot s+e/a\,|\,s\,|\,t\,\right),\;s,t\in \mathbb {R} \right\}}
oder etwas anders geschrieben
L
=
{
(
e
/
a
|
0
|
0
)
+
s
⋅
(
−
c
/
a
|
1
|
0
)
+
t
⋅
(
−
b
/
a
|
0
|
1
)
,
s
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{(\,e/a\,|\,0\,|\,0\,)+s\cdot (\,-c/a\,|\,1\,|\,0\,)+t\cdot (\,-b/a\,|\,0\,|\,1\,),\;s,t\in \mathbb {R} \right\}}
3
⋅
x
1
+
8
⋅
x
2
=
15
⇔
3
⋅
x
1
=
15
−
8
⋅
x
2
⇔
x
1
=
5
−
8
3
⋅
x
2
{\displaystyle {\begin{matrix}&3\cdot x_{1}&+&8\cdot x_{2}&=&15&&\\\Leftrightarrow &3\cdot x_{1}&&&=&15&-&8\cdot x_{2}\\\Leftrightarrow &x_{1}&&&=&5&-&{\frac {8}{3}}\cdot x_{2}\\\end{matrix}}}
L
=
{
(
5
−
8
3
⋅
t
|
t
)
,
t
∈
R
}
=
{
(
5
|
0
)
+
t
⋅
(
−
8
3
|
1
)
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5-{\frac {8}{3}}\cdot t\,|\,t\,\right),\;t\in \mathbb {R} \right\}=\left\{(\,5\,|\,0\,)+t\cdot (\,-{\frac {8}{3}}\,|\,1\,),\;t\in \mathbb {R} \right\}}
Setzt man
s
=
1
/
3
t
{\displaystyle s=1/3\,t}
t
=
3
s
{\displaystyle t=3\,s}
L
=
{
(
5
−
8
⋅
s
|
3
⋅
s
)
,
s
∈
R
}
=
{
(
5
|
0
)
+
s
⋅
(
−
8
|
3
)
,
s
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5-8\cdot s\,|\,3\cdot s\,\right),\;s\in \mathbb {R} \right\}=\left\{(\,5\,|\,0\,)+s\cdot (\,-8\,|\,3\,),\;s\in \mathbb {R} \right\}}
3
⋅
x
1
+
8
⋅
x
2
−
12
⋅
x
3
=
15
{\displaystyle 3\cdot x_{1}+8\cdot x_{2}-12\cdot x_{3}=15}
⇔
x
1
=
5
−
8
/
3
⋅
x
2
+
4
⋅
x
3
{\displaystyle \Leftrightarrow x_{1}=5-8/3\cdot x_{2}+4\cdot x_{3}}
Mit
x
2
=
3
s
{\displaystyle x_{2}=3\,s}
x
3
=
t
{\displaystyle x_{3}=t}
L
=
{
(
5
−
8
⋅
s
+
4
⋅
t
|
3
⋅
s
|
t
)
,
s
,
t
∈
R
}
=
{
(
5
|
0
|
0
)
+
s
⋅
(
−
8
|
3
|
0
)
+
t
⋅
(
4
|
0
|
1
)
,
s
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5-8\cdot s+4\cdot t\,|\,3\cdot s\,|\,t\,\right),\;s,t\in \mathbb {R} \right\}=\left\{(\,5\,|\,0\,|\,0\,)+s\cdot (\,-8\,|\,3\,|\,0\,)+t\cdot (\,4\,|\,0\,|\,1\,),\;s,t\in \mathbb {R} \right\}}
Ein solches Gleichungssystem sieht wie folgt aus:
a
1
⋅
x
1
+
b
1
⋅
x
2
+
c
1
⋅
x
3
=
e
1
a
2
⋅
x
1
+
b
2
⋅
x
2
+
c
2
⋅
x
3
=
e
2
{\displaystyle {\begin{array}{cccccccc}a_{1}\cdot x_{1}&+&b_{1}\cdot x_{2}&+&c_{1}\cdot x_{3}&=&e_{1}\\a_{2}\cdot x_{1}&+&b_{2}\cdot x_{2}&+&c_{2}\cdot x_{3}&=&e_{2}\\\end{array}}}
wobei nicht sowohl
a
1
{\displaystyle a_{1}}
a
2
{\displaystyle a_{2}}
b
1
{\displaystyle b_{1}}
b
2
{\displaystyle b_{2}}
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
a
1
≠
0
{\displaystyle a_{1}\neq 0}
a
1
⋅
x
1
+
b
1
⋅
x
2
+
c
1
⋅
x
3
=
e
1
n
2
⋅
x
2
+
k
2
⋅
x
3
=
l
2
{\displaystyle {\begin{array}{cccccccc}a_{1}\cdot x_{1}&+&b_{1}\cdot x_{2}&+&c_{1}\cdot x_{3}&=&e_{1}\\&&n_{2}\cdot x_{2}&+&k_{2}\cdot x_{3}&=&l_{2}\\\end{array}}}
An diesem Punkt müssen mehrere Fälle unterschieden werden:
1. Fall: Gilt
n
2
=
k
2
=
0
{\displaystyle n_{2}=k_{2}=0}
l
2
{\displaystyle l_{2}}
1.1. Fall: Gilt auch
l
2
=
0
{\displaystyle l_{2}=0}
0
=
0
{\displaystyle 0=0}
siehe hier )
1.2. Fall: Gilt dagegen
l
2
≠
0
{\displaystyle l_{2}\neq 0}
2. Fall: Wenigstens eine der beiden Zahlen
n
2
{\displaystyle n_{2}}
k
2
{\displaystyle k_{2}}
m
1
⋅
x
1
+
k
1
⋅
x
3
=
l
1
n
2
⋅
x
2
+
k
2
⋅
x
3
=
l
2
{\displaystyle {\begin{array}{|ccccccc|}m_{1}\cdot x_{1}&&&+&k_{1}\cdot x_{3}&=&l_{1}\\&&n_{2}\cdot x_{2}&+&k_{2}\cdot x_{3}&=&l_{2}\\\end{array}}}
m
1
⋅
x
1
+
n
1
⋅
x
2
=
l
1
n
2
⋅
x
2
+
k
2
⋅
x
3
=
l
2
{\displaystyle {\begin{array}{|ccccccc|}m_{1}\cdot x_{1}&+&n_{1}\cdot x_{2}&&&=&l_{1}\\&&n_{2}\cdot x_{2}&+&k_{2}\cdot x_{3}&=&l_{2}\\\end{array}}}
bzw. etwas anders geschrieben
x
1
=
−
k
1
/
m
1
⋅
x
3
+
l
1
/
m
1
x
2
=
−
k
2
/
n
2
⋅
x
3
+
l
2
/
n
2
{\displaystyle {\begin{array}{|cccccc|}x_{1}&=&-&k_{1}/m_{1}\cdot x_{3}&+&l_{1}/m_{1}\\x_{2}&=&-&k_{2}/n_{2}\cdot x_{3}&+&l_{2}/n_{2}\\\end{array}}}
x
1
=
−
n
1
/
m
1
⋅
x
2
+
l
1
/
m
1
x
3
=
−
n
2
/
k
2
⋅
x
2
+
l
2
/
k
2
{\displaystyle {\begin{array}{|cccccc|}x_{1}&=&-&n_{1}/m_{1}\cdot x_{2}&+&l_{1}/m_{1}\\x_{3}&=&-&n_{2}/k_{2}\cdot x_{2}&+&l_{2}/k_{2}\\\end{array}}}
Einmal lassen sich
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
3
{\displaystyle x_{3}}
x
1
{\displaystyle x_{1}}
x
3
{\displaystyle x_{3}}
x
2
{\displaystyle x_{2}}
f
1
=
−
k
1
/
m
1
{\displaystyle f_{1}=-k_{1}/m_{1}}
g
1
=
l
1
/
m
1
{\displaystyle g_{1}=l_{1}/m_{1}}
f
2
=
−
k
2
/
n
2
{\displaystyle f_{2}=-k_{2}/n_{2}}
g
2
=
l
2
/
n
2
{\displaystyle g_{2}=l_{2}/n_{2}}
f
1
=
−
n
1
/
m
1
{\displaystyle f_{1}=-n_{1}/m_{1}}
g
1
=
l
1
/
m
1
{\displaystyle g_{1}=l_{1}/m_{1}}
f
2
=
−
n
2
/
k
2
{\displaystyle f_{2}=-n_{2}/k_{2}}
g
1
=
l
2
/
k
2
{\displaystyle g_{1}=l_{2}/k_{2}}
x
1
=
f
1
⋅
x
3
+
g
1
x
2
=
f
2
⋅
x
3
+
g
2
{\displaystyle {\begin{array}{|cccc|}x_{1}&=f_{1}\cdot x_{3}&+&g_{1}\\x_{2}&=f_{2}\cdot x_{3}&+&g_{2}\\\end{array}}}
x
1
=
f
1
⋅
x
2
+
g
1
x
3
=
f
2
⋅
x
2
+
g
2
{\displaystyle {\begin{array}{|cccc|}x_{1}&=f_{1}\cdot x_{2}&+&g_{1}\\x_{3}&=f_{2}\cdot x_{2}&+&g_{2}\\\end{array}}}
Die Lösung des linearen Gleichungssystems sind demnach:
L
=
{
(
f
1
⋅
t
+
g
1
|
f
2
⋅
t
+
g
2
|
t
)
,
t
∈
R
}
=
{
(
g
1
|
g
2
|
0
)
+
t
⋅
(
f
1
|
f
2
|
1
)
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,f_{1}\cdot t+g_{1}\,|\,f_{2}\cdot t+g_{2}\,|\,t\,\right),\;t\in \mathbb {R} \right\}=\left\{(\,g_{1}\,|\,g_{2}\,|\,0\,)+t\cdot (\,f_{1}\,|\,f_{2}\,|\,1\,)\;t\in \mathbb {R} \right\}}
bzw.
L
=
{
(
f
1
⋅
t
+
g
1
|
t
|
f
2
⋅
t
+
g
2
)
,
t
∈
R
}
=
{
(
g
1
|
0
|
g
2
)
+
t
⋅
(
f
1
|
1
|
f
2
)
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,f_{1}\cdot t+g_{1}\,|\,t\,|\,f_{2}\cdot t+g_{2}\,\right),\;t\in \mathbb {R} \right\}=\left\{(\,g_{1}\,|\,0\,|\,g_{2}\,)+t\cdot (\,f_{1}\,|\,1\,|\,f_{2}\,)\;t\in \mathbb {R} \right\}}
(
1
)
3
⋅
x
1
+
7
⋅
x
2
−
1
⋅
x
3
=
24
(
2
)
−
6
⋅
x
1
−
5
⋅
x
2
+
20
⋅
x
3
=
33
{\displaystyle {\begin{array}{ccccccccc}(1)&3\cdot x_{1}&+&7\cdot x_{2}&-&1\cdot x_{3}&=&24\\(2)&-6\cdot x_{1}&-&5\cdot x_{2}&+&20\cdot x_{3}&=&33\\\end{array}}}
In schematischer Darstellung:
(
1.
)
(
2.
)
[
3
7
−
1
24
−
6
−
5
20
33
]
(
1.
)
⋅
2
(
2.
)
⇒
(
3.
)
(
4.
)
[
6
14
−
2
48
−
6
−
5
20
33
]
(
3.
)
:
2
(
3.
)
+
(
4.
)
{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\-6&-5&20&33\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 2\\(2.)\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}6&14&-2&48\\-6&-5&20&33\\\end{array}}\right]{\begin{array}{l}(3.):2\\(3.)+(4.)\\\end{array}}}
⇒
(
5.
)
(
6.
)
[
3
7
−
1
24
0
9
18
81
]
(
5.
)
(
6.
)
⋅
(
−
7
/
9
)
⇒
(
7.
)
(
8.
)
[
3
7
−
1
24
0
−
7
−
14
−
63
]
(
7.
)
+
(
8.
)
(
8.
)
:
(
−
7
)
{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\0&9&18&81\\\end{array}}\right]{\begin{array}{l}(5.)\\(6.)\cdot (-7/9)\\\end{array}}\Rightarrow {\begin{array}{c}(7.)\\(8.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\0&-7&-14&-63\\\end{array}}\right]{\begin{array}{l}(7.)+(8.)\\(8.):(-7)\\\end{array}}}
⇒
(
9.
)
(
10.
)
[
3
0
−
15
−
39
0
1
2
9
]
(
9.
)
:
3
(
10.
)
⇒
(
11.
)
(
12.
)
[
1
0
−
5
−
13
0
1
2
9
]
{\displaystyle \Rightarrow {\begin{array}{c}(9.)\\(10.)\\\end{array}}\left[{\begin{array}{ccc|c}3&0&-15&-39\\0&1&2&9\\\end{array}}\right]{\begin{array}{l}(9.):3\\(10.)\\\end{array}}\Rightarrow {\begin{array}{c}(11.)\\(12.)\\\end{array}}\left[{\begin{array}{ccc|c}1&0&-5&-13\\0&1&2&9\\\end{array}}\right]}
Wieder in der üblichen Schreibweise
1
⋅
x
1
−
5
⋅
x
3
=
−
13
1
⋅
x
2
+
2
⋅
x
3
=
9
⇒
x
1
=
5
⋅
x
3
−
13
x
2
=
−
2
⋅
x
3
+
9
{\displaystyle {\begin{array}{|ccccc|}1\cdot x_{1}&-&5\cdot x_{3}&=&-13\\1\cdot x_{2}&+&2\cdot x_{3}&=&9\\\end{array}}\Rightarrow {\begin{array}{|cccc|}x_{1}&=5\cdot x_{3}&-&13\\x_{2}&=-2\cdot x_{3}&+&9\\\end{array}}}
Die Lösungsmenge ist demnach:
L
=
{
(
5
⋅
t
−
13
|
−
2
⋅
t
+
9
|
t
)
,
t
∈
R
}
=
{
(
−
13
|
9
|
0
)
+
t
⋅
(
5
|
−
2
|
1
)
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5\cdot t-13\,|\,-2\cdot t+9\,|\,t\,\right),\;t\in \mathbb {R} \right\}=\left\{(\,-13\,|\,9\,|\,0\,)+t\cdot (\,5\,|\,-2\,|\,1\,)\;t\in \mathbb {R} \right\}}
(
1
)
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
(
2
)
−
21
⋅
x
1
−
56
⋅
x
2
+
84
⋅
x
3
=
−
100
{\displaystyle {\begin{array}{ccccccccc}(1)&6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\(2)&-21\cdot x_{1}&-&56\cdot x_{2}&+&84\cdot x_{3}&=&-100\\\end{array}}}
In schematischer Darstellung:
(
1.
)
(
2.
)
[
6
16
−
24
30
−
21
−
56
84
−
100
]
(
1.
)
⋅
7
(
2.
)
⋅
2
⇒
(
3.
)
(
4.
)
[
42
112
−
168
210
−
42
−
112
168
−
200
]
(
3.
)
(
3.
)
+
(
4.
)
{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&84&-100\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&168&-200\\\end{array}}\right]{\begin{array}{l}(3.)\\(3.)+(4.)\\\end{array}}}
⇒
(
5.
)
(
6.
)
[
42
112
−
168
210
0
0
0
10
]
{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\0&0&0&10\\\end{array}}\right]}
Dieses Gleichungssystem hat wegen der falschen Aussage in der letzten Zeile (0=10) keine Lösung. Die Lösungsmenge ist demnach:
L
=
∅
{\displaystyle {\mathfrak {L}}=\emptyset }
(
1
)
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
(
2
)
−
21
⋅
x
1
−
56
⋅
x
2
+
84
⋅
x
3
=
−
105
{\displaystyle {\begin{array}{ccccccccc}(1)&6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\(2)&-21\cdot x_{1}&-&56\cdot x_{2}&+&84\cdot x_{3}&=&-105\\\end{array}}}
In schematischer Darstellung:
(
1.
)
(
2.
)
[
6
16
−
24
30
−
21
−
56
84
−
105
]
(
1.
)
⋅
7
(
2.
)
⋅
2
⇒
(
3.
)
(
4.
)
[
42
112
−
168
210
−
42
−
112
168
−
210
]
(
3.
)
:
14
(
3.
)
+
(
4.
)
{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&84&-105\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&168&-210\\\end{array}}\right]{\begin{array}{l}(3.):14\\(3.)+(4.)\\\end{array}}}
⇒
(
5.
)
(
6.
)
[
3
8
−
12
15
0
0
0
0
]
{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&0&0\\\end{array}}\right]}
Dieses Gleichungssystem hat nur eine für die Lösung relevante Gleichung. Die Lösungsmenge ist (siehe Beispiele zu einer Gleichung):
L
=
{
(
5
−
8
⋅
s
+
4
⋅
t
|
3
⋅
s
|
t
)
,
s
,
t
∈
R
}
=
{
(
5
|
0
|
0
)
+
s
⋅
(
−
8
|
3
|
0
)
+
t
⋅
(
4
|
0
|
1
)
,
s
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5-8\cdot s+4\cdot t\,|\,3\cdot s\,|\,t\,\right),\;s,t\in \mathbb {R} \right\}=\left\{(\,5\,|\,0\,|\,0\,)+s\cdot (\,-8\,|\,3\,|\,0\,)+t\cdot (\,4\,|\,0\,|\,1\,),\;s,t\in \mathbb {R} \right\}}
(
1
)
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
(
2
)
−
21
⋅
x
1
−
56
⋅
x
2
+
85
⋅
x
3
=
−
100
{\displaystyle {\begin{array}{ccccccccc}(1)&6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\(2)&-21\cdot x_{1}&-&56\cdot x_{2}&+&85\cdot x_{3}&=&-100\\\end{array}}}
In schematischer Darstellung:
(
1.
)
(
2.
)
[
6
16
−
24
30
−
21
−
56
85
−
100
]
(
1.
)
⋅
7
(
2.
)
⋅
2
⇒
(
3.
)
(
4.
)
[
42
112
−
168
210
−
42
−
112
170
−
200
]
(
3.
)
:
14
(
3.
)
+
(
4.
)
{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&85&-100\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&170&-200\\\end{array}}\right]{\begin{array}{l}(3.):14\\(3.)+(4.)\\\end{array}}}
⇒
(
5.
)
(
6.
)
[
3
8
−
12
15
0
0
2
10
]
(
5.
)
(
6.
)
⋅
6
⇒
(
7.
)
(
8.
)
[
3
8
−
12
15
0
0
12
60
]
(
7.
)
+
(
8.
)
(
8.
)
:
12
{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&2&10\\\end{array}}\right]{\begin{array}{l}(5.)\\(6.)\cdot 6\\\end{array}}\Rightarrow {\begin{array}{c}(7.)\\(8.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&12&60\\\end{array}}\right]{\begin{array}{l}(7.)+(8.)\\(8.):12\\\end{array}}}
⇒
(
9.
)
(
10.
)
[
3
8
0
75
0
0
1
5
]
{\displaystyle \Rightarrow {\begin{array}{c}(9.)\\(10.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&0&75\\0&0&1&5\\\end{array}}\right]}
Wieder in der üblichen Schreibweise
3
⋅
x
1
+
8
⋅
x
2
=
75
x
3
=
5
⇒
x
1
=
−
8
/
3
⋅
x
2
+
25
x
3
=
5
{\displaystyle {\begin{array}{|ccccc|}3\cdot x_{1}&+&8\cdot x_{2}&=&75\\&&x_{3}&=&5\\\end{array}}\Rightarrow {\begin{array}{|ccccc|}x_{1}&=&-8/3\cdot x_{2}&+&25\\x_{3}&=&5&&\\\end{array}}}
Die Lösungsmenge ist demnach
L
=
{
(
25
−
8
/
3
⋅
t
|
t
|
5
)
,
t
∈
R
}
=
{
(
25
|
0
|
5
)
+
t
⋅
(
−
8
/
3
|
1
|
0
)
,
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,25-8/3\cdot t\,|\,t\,|\,5\,\right),t\in \mathbb {R} \right\}=\left\{(\,25\,|\,0\,|\,5\,)+t\cdot (\,-8/3\,|\,1\,|\,0\,),\;t\in \mathbb {R} \right\}}
oder wenn man Brüche lieber vermeiden will, mit
s
=
1
/
3
⋅
t
{\displaystyle s=1/3\cdot t}
L
=
{
(
25
−
8
⋅
s
|
3
⋅
s
|
5
)
,
s
∈
R
}
=
{
(
25
|
0
|
5
)
+
s
⋅
(
−
8
|
3
|
0
)
,
s
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,25-8\cdot s\,|\,3\cdot s\,|\,5\,\right),s\in \mathbb {R} \right\}=\left\{(\,25\,|\,0\,|\,5\,)+s\cdot (\,-8\,|\,3\,|\,0\,),\;s\in \mathbb {R} \right\}}
2 Gleichungen und 3 Unbekannte mit dem Determinanten-Verfahren
Bearbeiten
Bei zwei Gleichungen mit 3 Unbekannten besteht die Koeffizientenmatrix aus 2 Zeilen und 3 Spalten. Determinanten kann man aber nur von quadratischen Matrizen (= gleich viele Zeilen und Spalten) berechnen. Trotzdem kann man mit einem kleinen Trick auch hier das Determinanten-Verfahren anwenden. Dazu schreibt man das Gleichungssystem
a
1
⋅
x
1
+
b
1
⋅
x
2
+
c
1
⋅
x
3
=
e
1
a
2
⋅
x
1
+
b
2
⋅
x
2
+
c
2
⋅
x
3
=
e
2
{\displaystyle {\begin{array}{cccccccc}a_{1}\cdot x_{1}&+&b_{1}\cdot x_{2}&+&c_{1}\cdot x_{3}&=&e_{1}\\a_{2}\cdot x_{1}&+&b_{2}\cdot x_{2}&+&c_{2}\cdot x_{3}&=&e_{2}\\\end{array}}}
um und erhält mit
x
3
=
t
{\displaystyle x_{3}=t}
a
1
⋅
x
1
+
b
1
⋅
x
2
=
e
1
−
c
1
⋅
t
a
2
⋅
x
1
+
b
2
⋅
x
2
=
e
2
−
c
2
⋅
t
{\displaystyle {\begin{array}{cccccccc}a_{1}\cdot x_{1}&+&b_{1}\cdot x_{2}&=&e_{1}&-&c_{1}\cdot t\\a_{2}\cdot x_{1}&+&b_{2}\cdot x_{2}&=&e_{2}&-&c_{2}\cdot t\\\end{array}}}
Die für das Determinantenverfahren wichtigen Matrizen sind dann:
A
=
(
a
1
b
1
a
2
b
2
)
{\displaystyle A=\left({\begin{array}{cc}a_{1}&b_{1}\\a_{2}&b_{2}\\\end{array}}\right)}
A
1
(
t
)
=
(
e
1
−
c
1
t
b
1
e
2
−
c
2
t
b
2
)
{\displaystyle A_{1}(t)=\left({\begin{array}{cc}e_{1}-c_{1}\,t&b_{1}\\e_{2}-c_{2}\,t&b_{2}\\\end{array}}\right)}
A
2
(
t
)
=
(
a
1
e
1
−
c
1
t
a
2
e
2
−
c
2
t
)
{\displaystyle A_{2}(t)=\left({\begin{array}{cc}a_{1}&e_{1}-c_{1}\,t\\a_{2}&e_{2}-c_{2}\,t\\\end{array}}\right)}
Falls
det
(
A
)
=
a
1
⋅
b
2
−
a
2
⋅
b
1
≠
0
{\displaystyle \det(A)=a_{1}\cdot b_{2}-a_{2}\cdot b_{1}\neq 0}
x
1
=
det
(
A
1
(
t
)
)
det
(
A
)
,
x
2
=
det
(
A
2
(
t
)
)
det
(
A
)
,
x
3
=
t
{\displaystyle x_{1}={\frac {\det \left(A_{1}(t)\right)}{\det(A)}},\;x_{2}={\frac {\det \left(A_{2}(t)\right)}{\det(A)}},\;x_{3}=t}
Ist dagegen
det
(
A
)
=
0
{\displaystyle \det(A)=0}
x
3
{\displaystyle x_{3}}
x
2
{\displaystyle x_{2}}
x
1
{\displaystyle x_{1}}
det
(
A
′
)
=
det
(
a
1
c
1
a
2
c
2
)
≠
0
{\displaystyle \det(A')=\det \left({\begin{array}{cc}a_{1}&c_{1}\\a_{2}&c_{2}\\\end{array}}\right)\neq 0}
A
1
′
(
t
)
=
(
e
1
−
b
1
t
c
1
e
2
−
b
2
t
c
2
)
{\displaystyle A_{1}'(t)=\left({\begin{array}{cc}e_{1}-b_{1}\,t&c_{1}\\e_{2}-b_{2}\,t&c_{2}\\\end{array}}\right)}
A
3
′
(
t
)
=
(
a
1
e
1
−
b
1
t
a
2
e
2
−
b
2
t
)
{\displaystyle A_{3}'(t)=\left({\begin{array}{cc}a_{1}&e_{1}-b_{1}\,t\\a_{2}&e_{2}-b_{2}\,t\\\end{array}}\right)}
Die Lösung bestimmen:
x
1
=
det
(
A
1
′
(
t
)
)
det
(
A
)
,
x
2
=
t
,
x
3
=
det
(
A
3
′
(
t
)
)
det
(
A
)
{\displaystyle x_{1}={\frac {\det \left(A_{1}'(t)\right)}{\det(A)}},\;x_{2}=t,\;x_{3}={\frac {\det \left(A_{3}'(t)\right)}{\det(A)}}}
Selbst wenn für alle drei Koeffizientenmatrizen gilt
det
(
A
)
=
det
(
A
′
)
=
det
(
A
″
)
=
0
{\displaystyle \det(A)=\det(A')=\det(A'')=0}
det
(
A
)
=
det
(
A
′
)
=
det
(
A
″
)
=
0
{\displaystyle \det(A)=\det(A')=\det(A'')=0}
det
(
B
)
=
det
(
a
1
e
1
a
2
e
2
)
{\displaystyle \det(B)=\det \left({\begin{array}{cc}a_{1}&e_{1}\\a_{2}&e_{2}\\\end{array}}\right)}
Gilt
det
(
A
)
=
det
(
A
′
)
=
det
(
A
″
)
=
det
(
B
)
=
0
{\displaystyle \det(A)=\det(A')=\det(A'')=\det(B)=0}
Gilt
det
(
A
)
=
det
(
A
′
)
=
det
(
A
″
)
=
0
{\displaystyle \det(A)=\det(A')=\det(A'')=0}
det
(
B
)
≠
0
{\displaystyle \det(B)\neq 0}
3
⋅
x
1
+
7
⋅
x
2
−
1
⋅
x
3
=
24
−
6
⋅
x
1
−
5
⋅
x
2
+
20
⋅
x
3
=
33
{\displaystyle {\begin{array}{cccccccc}3\cdot x_{1}&+&7\cdot x_{2}&-&1\cdot x_{3}&=&24\\-6\cdot x_{1}&-&5\cdot x_{2}&+&20\cdot x_{3}&=&33\\\end{array}}}
x
3
=
t
{\displaystyle x_{3}=t}
3
⋅
x
1
+
7
⋅
x
2
=
24
+
1
⋅
t
−
6
⋅
x
1
−
5
⋅
x
2
=
33
−
20
⋅
t
{\displaystyle {\begin{array}{cccccccc}3\cdot x_{1}&+&7\cdot x_{2}&=&24&+&1\cdot t\\-6\cdot x_{1}&-&5\cdot x_{2}&=&33&-&20\cdot t\\\end{array}}}
Die für das Determinantenverfahren wichtigen Determinanten sind dann:
det
(
A
)
=
det
(
3
7
−
6
−
5
)
=
27
{\displaystyle \det(A)=\det \left({\begin{array}{cc}3&7\\-6&-5\\\end{array}}\right)=27}
det
(
A
1
(
t
)
)
=
det
(
24
+
t
7
33
−
20
t
−
5
)
=
−
5
⋅
(
24
+
t
)
−
7
⋅
(
33
−
20
t
)
=
135
t
−
351
{\displaystyle \det(A_{1}(t))=\det \left({\begin{array}{cc}24+t&7\\33-20\,t&-5\\\end{array}}\right)=-5\cdot (24+t)-7\cdot (33-20\,t)=135\,t-351}
det
(
A
2
(
t
)
)
=
det
(
3
24
+
t
−
6
33
−
20
t
)
=
3
⋅
(
33
−
20
t
)
+
6
⋅
(
24
+
t
)
=
−
54
t
+
243
{\displaystyle \det(A_{2}(t))=\det \left({\begin{array}{cc}3&24+t\\-6&33-20\,t\\\end{array}}\right)=3\cdot (33-20\,t)+6\cdot (24+t)=-54\,t+243}
Wegen
det
(
A
)
≠
0
{\displaystyle \det(A)\neq 0}
x
1
=
135
t
−
351
27
=
5
t
−
13
,
x
2
=
−
54
t
+
243
27
=
−
2
t
+
9
,
x
3
=
t
{\displaystyle x_{1}={\frac {135\,t-351}{27}}=5\,t-13,\;x_{2}={\frac {-54\,t+243}{27}}=-2\,t+9,\;x_{3}=t}
Die Lösungsmenge ist also:
L
=
{
(
5
⋅
t
−
13
|
−
2
⋅
t
+
9
|
t
)
,
t
∈
R
}
=
{
(
−
13
|
9
|
0
)
+
t
⋅
(
5
|
−
2
|
1
)
t
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,5\cdot t-13\,|\,-2\cdot t+9\,|\,t\,\right),\;t\in \mathbb {R} \right\}=\left\{(\,-13\,|\,9\,|\,0\,)+t\cdot (\,5\,|\,-2\,|\,1\,)\;t\in \mathbb {R} \right\}}
(
1
)
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
(
2
)
−
21
⋅
x
1
−
56
⋅
x
2
+
85
⋅
x
3
=
−
100
{\displaystyle {\begin{array}{ccccccccc}(1)&6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\(2)&-21\cdot x_{1}&-&56\cdot x_{2}&+&85\cdot x_{3}&=&-100\\\end{array}}}
x
3
=
t
{\displaystyle x_{3}=t}
6
⋅
x
1
+
16
⋅
x
2
=
30
+
24
⋅
t
−
21
⋅
x
1
−
56
⋅
x
2
=
−
100
−
85
⋅
t
{\displaystyle {\begin{array}{cccccccc}6\cdot x_{1}&+&16\cdot x_{2}&=&30&+&24\cdot t\\-21\cdot x_{1}&-&56\cdot x_{2}&=&-100&-&85\cdot t\\\end{array}}}
Die Koeffizientenmatrix dazu hat die Determinante:
det
(
A
)
=
det
(
6
16
−
21
−
56
)
=
0
{\displaystyle \det(A)=\det \left({\begin{array}{cc}6&16\\-21&-56\\\end{array}}\right)=0}
hilft uns also nicht weiter. Zweiter Versuch
x
2
=
s
{\displaystyle x_{2}=s}
6
⋅
x
1
−
24
⋅
x
3
=
30
−
16
⋅
s
−
21
⋅
x
1
+
85
⋅
x
3
=
−
100
+
56
⋅
s
{\displaystyle {\begin{array}{cccccccc}6\cdot x_{1}&-&24\cdot x_{3}&=&30&-&16\cdot s\\-21\cdot x_{1}&+&85\cdot x_{3}&=&-100&+&56\cdot s\\\end{array}}}
Für die entsprechende Koeffizientenmatrix gilt:
det
(
A
′
)
=
det
(
6
−
24
−
21
85
)
=
6
{\displaystyle \det(A')=\det \left({\begin{array}{cc}6&-24\\-21&85\\\end{array}}\right)=6}
Wegen
det
(
A
′
(
s
)
)
≠
0
{\displaystyle \det(A'(s))\neq 0}
det
(
A
1
′
(
s
)
)
=
det
(
30
−
16
s
−
24
−
100
+
56
s
85
)
=
85
⋅
(
30
−
16
s
)
+
24
⋅
(
−
100
+
56
s
)
=
−
16
s
+
150
{\displaystyle \det(A_{1}'(s))=\det \left({\begin{array}{cc}30-16\,s&-24\\-100+56\,s&85\\\end{array}}\right)=85\cdot (30-16\,s)+24\cdot (-100+56\,s)=-16\,s+150}
det
(
A
3
′
(
s
)
)
=
det
(
6
30
−
16
s
−
21
−
100
+
56
s
)
=
6
⋅
(
−
100
+
56
s
)
+
21
⋅
(
30
−
16
s
)
=
30
{\displaystyle \det(A_{3}'(s))=\det \left({\begin{array}{cc}6&30-16\,s\\-21&-100+56\,s\\\end{array}}\right)=6\cdot (-100+56\,s)+21\cdot (30-16\,s)=30}
Es gilt:
x
1
=
−
16
s
+
150
6
=
−
8
3
s
+
25
,
x
2
=
s
,
x
3
=
30
6
=
5
{\displaystyle x_{1}={\frac {-16\,s+150}{6}}=-{\frac {8}{3}}\,s+25,\;x_{2}=s,\;x_{3}={\frac {30}{6}}=5}
Die Lösungsmenge ist demnach
L
=
{
(
25
−
8
/
3
⋅
s
|
s
|
5
)
,
s
∈
R
}
=
{
(
25
|
0
|
5
)
+
s
⋅
(
−
8
/
3
|
1
|
0
)
,
s
∈
R
}
{\displaystyle {\mathfrak {L}}=\left\{\left(\,25-8/3\cdot s\,|\,s\,|\,5\,\right),s\in \mathbb {R} \right\}=\left\{(\,25\,|\,0\,|\,5\,)+s\cdot (\,-8/3\,|\,1\,|\,0\,),\;s\in \mathbb {R} \right\}}
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
−
21
⋅
x
1
−
56
⋅
x
2
+
84
⋅
x
3
=
−
100
{\displaystyle {\begin{array}{cccccccc}6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\-21\cdot x_{1}&-&56\cdot x_{2}&+&84\cdot x_{3}&=&-100\\\end{array}}}
x
3
=
t
{\displaystyle x_{3}=t}
6
⋅
x
1
+
16
⋅
x
2
=
30
+
24
⋅
t
−
21
⋅
x
1
−
56
⋅
x
2
=
−
100
−
84
⋅
t
{\displaystyle {\begin{array}{cccccccc}6\cdot x_{1}&+&16\cdot x_{2}&=&30&+&24\cdot t\\-21\cdot x_{1}&-&56\cdot x_{2}&=&-100&-&84\cdot t\\\end{array}}}
Die Koeffizientenmatrix dazu hat die Determinante:
det
(
A
)
=
det
(
6
16
−
21
−
56
)
=
0
{\displaystyle \det(A)=\det \left({\begin{array}{cc}6&16\\-21&-56\\\end{array}}\right)=0}
hilft uns also nicht weiter. Zweiter Versuch
x
2
=
s
{\displaystyle x_{2}=s}
6
⋅
x
1
−
24
⋅
x
3
=
30
−
16
⋅
s
−
21
⋅
x
1
+
84
⋅
x
3
=
−
100
+
56
⋅
s
{\displaystyle {\begin{array}{cccccccc}6\cdot x_{1}&-&24\cdot x_{3}&=&30&-&16\cdot s\\-21\cdot x_{1}&+&84\cdot x_{3}&=&-100&+&56\cdot s\\\end{array}}}
Für die entsprechende Koeffizientenmatrix gilt:
det
(
A
′
)
=
det
(
6
−
24
−
21
84
)
=
0
{\displaystyle \det(A')=\det \left({\begin{array}{cc}6&-24\\-21&84\\\end{array}}\right)=0}
Als letzter Versuch
x
1
=
r
{\displaystyle x_{1}=r}
16
⋅
x
2
−
24
⋅
x
3
=
30
−
6
⋅
r
−
56
⋅
x
2
+
84
⋅
x
3
=
−
100
+
21
⋅
r
{\displaystyle {\begin{array}{cccccccc}16\cdot x_{2}&-&24\cdot x_{3}&=&30&-&6\cdot r\\-56\cdot x_{2}&+&84\cdot x_{3}&=&-100&+&21\cdot r\\\end{array}}}
aber auch für die hier auftretende Matrix gilt:
det
(
A
″
)
=
det
(
16
−
24
−
56
84
)
=
0
{\displaystyle \det(A'')=\det \left({\begin{array}{cc}16&-24\\-56&84\\\end{array}}\right)=0}
Entscheident ist also die Matrix B.
det
(
B
)
=
det
(
6
30
−
21
−
100
)
=
30
≠
0
{\displaystyle \det(B)=\det \left({\begin{array}{cc}6&30\\-21&-100\\\end{array}}\right)=30\neq 0}
Also besitzt das Gleichungssystem keine Lösung
L
=
∅
{\displaystyle {\mathfrak {L}}=\emptyset }
(
1
)
6
⋅
x
1
+
16
⋅
x
2
−
24
⋅
x
3
=
30
(
2
)
−
21
⋅
x
1
−
56
⋅
x
2
+
84
⋅
x
3
=
−
105
{\displaystyle {\begin{array}{ccccccccc}(1)&6\cdot x_{1}&+&16\cdot x_{2}&-&24\cdot x_{3}&=&30\\(2)&-21\cdot x_{1}&-&56\cdot x_{2}&+&84\cdot x_{3}&=&-105\\\end{array}}}
Auch hier gilt natürlich:
det
(
A
)
=
det
(
A
′
)
=
det
(
A
″
)
=
0
{\displaystyle \det(A)=\det(A')=\det(A'')=0}
Aber im Gegensatz zum 3. Beispiel gilt hier:
det
(
B
)
=
det
(
6
30
−
21
−
105
)
=
0
{\displaystyle \det(B)=\det \left({\begin{array}{cc}6&30\\-21&-105\\\end{array}}\right)=0}
Das liefert uns zwar keine Lösung des Gleichungssystems, aber immerhin wissen wir jetzt, dass es Lösungen gibt.
![{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\-6&-5&20&33\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 2\\(2.)\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}6&14&-2&48\\-6&-5&20&33\\\end{array}}\right]{\begin{array}{l}(3.):2\\(3.)+(4.)\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bef09de295efdaea13ce16a9b961fa17a96f7516)
![{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\0&9&18&81\\\end{array}}\right]{\begin{array}{l}(5.)\\(6.)\cdot (-7/9)\\\end{array}}\Rightarrow {\begin{array}{c}(7.)\\(8.)\\\end{array}}\left[{\begin{array}{ccc|c}3&7&-1&24\\0&-7&-14&-63\\\end{array}}\right]{\begin{array}{l}(7.)+(8.)\\(8.):(-7)\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67fd8c493c359ad108423fea95c63bee0a7aa6c5)
![{\displaystyle \Rightarrow {\begin{array}{c}(9.)\\(10.)\\\end{array}}\left[{\begin{array}{ccc|c}3&0&-15&-39\\0&1&2&9\\\end{array}}\right]{\begin{array}{l}(9.):3\\(10.)\\\end{array}}\Rightarrow {\begin{array}{c}(11.)\\(12.)\\\end{array}}\left[{\begin{array}{ccc|c}1&0&-5&-13\\0&1&2&9\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac43f66c22e52927b4b57b78596cd450fb28890b)
![{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&84&-100\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&168&-200\\\end{array}}\right]{\begin{array}{l}(3.)\\(3.)+(4.)\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfe709308bf05bd0d912d83218a719826f6630ad)
![{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\0&0&0&10\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14d6a5cb80048f0f3056a9727eb9a5de8351d74b)
![{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&84&-105\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&168&-210\\\end{array}}\right]{\begin{array}{l}(3.):14\\(3.)+(4.)\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69b5fda328377480bff1a23a4714ed8d1641e8d6)
![{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&0&0\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c756f44b2d4a0fad920a8d959110946d3e2887b4)
![{\displaystyle {\begin{array}{c}(1.)\\(2.)\\\end{array}}\left[{\begin{array}{ccc|c}6&16&-24&30\\-21&-56&85&-100\\\end{array}}\right]{\begin{array}{l}(1.)\cdot 7\\(2.)\cdot 2\\\end{array}}\Rightarrow {\begin{array}{c}(3.)\\(4.)\\\end{array}}\left[{\begin{array}{ccc|c}42&112&-168&210\\-42&-112&170&-200\\\end{array}}\right]{\begin{array}{l}(3.):14\\(3.)+(4.)\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beb20e1d64e4d14ce195df0960a16e659c6ff07b)
![{\displaystyle \Rightarrow {\begin{array}{c}(5.)\\(6.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&2&10\\\end{array}}\right]{\begin{array}{l}(5.)\\(6.)\cdot 6\\\end{array}}\Rightarrow {\begin{array}{c}(7.)\\(8.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&-12&15\\0&0&12&60\\\end{array}}\right]{\begin{array}{l}(7.)+(8.)\\(8.):12\\\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4af9b3132226a70d48cf43c6f2b1df0219a42db1)
![{\displaystyle \Rightarrow {\begin{array}{c}(9.)\\(10.)\\\end{array}}\left[{\begin{array}{ccc|c}3&8&0&75\\0&0&1&5\\\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecef1c253d87c6ab55bf9e1dc9c01f3a5290cc9a)
