Problem 1
Decide if the vector is a member of the set.
(
2
3
)
{\displaystyle {\begin{pmatrix}2\\3\end{pmatrix}}}
{
(
1
2
)
k
|
k
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}
(
−
3
3
)
{\displaystyle {\begin{pmatrix}-3\\3\end{pmatrix}}}
{
(
1
−
1
)
k
|
k
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\-1\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}
(
−
3
3
4
)
{\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}
{
(
1
−
1
2
)
k
|
k
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}
(
−
3
3
4
)
{\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}
{
(
1
−
1
2
)
k
+
(
0
0
2
)
m
|
k
,
m
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k+{\begin{pmatrix}0\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}
(
1
4
14
)
{\displaystyle {\begin{pmatrix}1\\4\\14\end{pmatrix}}}
{
(
2
2
5
)
k
+
(
−
1
0
2
)
m
|
k
,
m
∈
R
}
{\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}
(
1
4
6
)
{\displaystyle {\begin{pmatrix}1\\4\\6\end{pmatrix}}}
{
(
2
2
5
)
k
+
(
−
1
0
2
)
m
|
k
,
m
∈
R
}
{\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}
Answer
No.
Yes.
No.
Yes.
Yes; use Gauss' method to get
k
=
4
{\displaystyle k=4}
m
=
−
3
{\displaystyle m=-3}
No; use Gauss' method to conclude that there is no solution.
Problem 2
Produce two descriptions of this set that are different than this one.
{
(
2
−
5
)
k
|
k
∈
R
}
{\displaystyle \{{\begin{pmatrix}2\\-5\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}
Answer
One easy thing to do is to double and triple the vector:
{
(
4
−
10
)
k
|
k
∈
R
}
and
{
(
6
−
15
)
k
|
k
∈
R
}
.
{\displaystyle \{{\begin{pmatrix}4\\-10\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}6\\-15\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}.}
This exercise is recommended for all readers.
Problem 3
Show that the three descriptions given at the start of this
subsection all describe the same set.
Answer
Instead of showing all three equalities, we can show that the first
equals the second, and that the second equals the third.
Both equalities are easy, using the methods of this subsection.
This exercise is recommended for all readers.
Problem 4
Show that these sets are equal
{
(
1
4
1
1
)
+
(
−
1
0
1
0
)
z
|
z
∈
R
}
and
{
(
0
4
2
1
)
+
(
−
1
0
1
0
)
k
|
k
∈
R
}
,
{\displaystyle \{{\begin{pmatrix}1\\4\\1\\1\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}0\\4\\2\\1\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \},}
and that both describe the solution set of this system.
x
−
y
+
z
+
w
=
−
1
y
−
w
=
3
x
+
z
+
2
w
=
4
{\displaystyle {\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\x&&&+&z&+&2w&=&4\end{array}}}
Answer
That system reduces like this:
→
−
ρ
1
+
ρ
2
x
−
y
+
z
+
w
=
−
1
y
−
w
=
3
y
+
w
=
5
→
−
ρ
2
+
ρ
3
x
−
y
+
z
+
w
=
−
1
y
−
w
=
3
2
w
=
2
{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\&&y&&&+&w&=&5\end{array}}\\&{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}&{\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\&&&&&&2w&=&2\end{array}}\end{array}}}
showing that
w
=
1
{\displaystyle w=1}
y
=
4
{\displaystyle y=4}
x
=
2
−
z
{\displaystyle x=2-z}
This exercise is recommended for all readers.
Problem 5
Decide if the sets are equal.
{
(
1
2
)
+
(
0
3
)
t
|
t
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}}
{
(
1
8
)
+
(
0
−
1
)
s
|
s
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s\,{\big |}\,s\in \mathbb {R} \}}
{
(
1
3
1
)
t
+
(
2
1
5
)
s
|
t
,
s
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\1\\5\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}
{
(
4
7
7
)
m
+
(
−
4
−
2
−
10
)
n
|
m
,
n
∈
R
}
{\displaystyle \{{\begin{pmatrix}4\\7\\7\end{pmatrix}}m+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}
{
(
1
2
)
t
|
t
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}}
{
(
2
4
)
m
+
(
4
8
)
n
|
m
,
n
∈
R
}
{\displaystyle \{{\begin{pmatrix}2\\4\end{pmatrix}}m+{\begin{pmatrix}4\\8\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}
{
(
1
0
2
)
s
+
(
−
1
1
0
)
t
|
s
,
t
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\0\\2\end{pmatrix}}s+{\begin{pmatrix}-1\\1\\0\end{pmatrix}}t\,{\big |}\,s,t\in \mathbb {R} \}}
{
(
−
1
1
1
)
m
+
(
0
1
3
)
n
|
m
,
n
∈
R
}
{\displaystyle \{{\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}
{
(
1
3
1
)
t
+
(
2
4
6
)
s
|
t
,
s
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\4\\6\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}
{
(
3
7
7
)
t
+
(
1
3
1
)
s
|
t
,
s
∈
R
}
{\displaystyle \{{\begin{pmatrix}3\\7\\7\end{pmatrix}}t+{\begin{pmatrix}1\\3\\1\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}
Answer
For each item, we call the first set
S
1
{\displaystyle S_{1}}
S
2
{\displaystyle S_{2}}
They are equal.
To see that
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
v
→
=
(
1
2
)
+
(
0
3
)
t
{\displaystyle {\vec {v}}={\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t}
s
{\displaystyle s}
v
→
=
(
1
8
)
+
(
0
−
1
)
s
.
{\displaystyle {\vec {v}}={\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s.}
t
{\displaystyle t}
s
{\displaystyle s}
1
+
0
s
=
1
+
0
t
8
−
1
s
=
2
+
3
t
{\displaystyle {\begin{array}{*{2}{rc}r}1&+&0s&=&1+0t\\8&-&1s&=&2+3t\end{array}}}
1
=
1
s
=
6
−
3
t
{\displaystyle {\begin{array}{*{1}{rc}r}1&=&1\\s&=&6-3t\end{array}}}
(
1
2
)
+
(
0
3
)
t
=
(
1
8
)
+
(
0
−
1
)
(
6
−
3
t
)
{\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t={\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}(6-3t)}
S
1
{\displaystyle S_{1}}
S
2
{\displaystyle S_{2}}
S
2
⊆
S
1
{\displaystyle S_{2}\subseteq S_{1}}
t
{\displaystyle t}
1
+
0
t
=
1
+
0
s
2
+
3
t
=
8
−
1
s
{\displaystyle {\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\2&+&3t&=&8-1s\end{array}}}
1
=
1
t
=
2
−
(
1
/
3
)
s
{\displaystyle {\begin{array}{*{1}{rc}r}1&=&1\\t&=&2-(1/3)s\end{array}}}
(
1
8
)
+
(
0
−
1
)
s
=
(
1
2
)
+
(
0
3
)
(
2
−
(
1
/
3
)
s
)
.
{\displaystyle {\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s={\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}(2-(1/3)s).}
These two are equal.
To show that
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
t
,
s
{\displaystyle t,s}
m
,
n
{\displaystyle m,n}
4
m
−
4
n
=
1
t
+
2
s
7
m
−
2
n
=
3
t
+
1
s
7
m
−
10
n
=
1
t
+
5
s
{\displaystyle {\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\7m&-&2n&=&3t+1s\\7m&-&10n&=&1t+5s\end{array}}}
(
4
−
4
1
t
+
2
s
7
−
2
3
t
+
1
s
7
−
10
1
t
+
5
s
)
→
(
−
7
/
4
)
ρ
1
+
ρ
3
(
−
7
/
4
)
ρ
1
+
ρ
2
(
4
−
4
1
t
+
2
s
0
5
(
5
/
4
)
t
−
(
10
/
4
)
s
0
−
3
−
(
3
/
4
)
t
+
(
6
/
4
)
s
)
→
(
3
/
5
)
ρ
2
+
ρ
3
(
4
−
4
1
t
+
2
s
0
5
(
5
/
4
)
t
−
(
10
/
4
)
s
0
0
0
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\7&-2&3t+1s\\7&-10&1t+5s\end{array}}\right)&{\xrightarrow[{(-7/4)\rho _{1}+\rho _{3}}]{(-7/4)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&-3&-(3/4)t+(6/4)s\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&0&0\end{array}}\right)\end{array}}}
(
1
3
1
)
t
+
(
2
1
5
)
s
=
(
4
7
7
)
(
(
1
/
2
)
t
)
+
(
−
4
−
2
−
10
)
(
(
1
/
4
)
t
−
(
1
/
2
)
s
)
{\displaystyle {\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\1\\5\end{pmatrix}}s={\begin{pmatrix}4\\7\\7\end{pmatrix}}((1/2)t)+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}((1/4)t-(1/2)s)}
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
S
2
⊆
S
1
{\displaystyle S_{2}\subseteq S_{1}}
1
t
+
2
s
=
4
m
−
4
n
3
t
+
1
s
=
7
m
−
2
n
1
t
+
5
s
=
7
m
−
10
n
{\displaystyle {\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\3t&+&1s&=&7m-2n\\1t&+&5s&=&7m-10n\end{array}}}
(
1
2
4
m
−
4
n
3
1
7
m
−
2
n
1
5
7
m
−
10
n
)
→
−
ρ
1
+
ρ
3
−
3
ρ
1
+
ρ
2
(
1
2
4
m
−
4
n
0
−
5
−
5
m
+
10
n
0
3
3
m
−
6
n
)
→
(
3
/
5
)
ρ
2
+
ρ
3
(
1
2
4
m
−
4
n
0
−
5
−
5
m
+
10
n
0
0
0
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\3&1&7m-2n\\1&5&7m-10n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&3&3m-6n\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&0&0\end{array}}\right)\end{array}}}
(
4
7
7
)
m
+
(
−
4
−
2
−
10
)
n
=
(
1
3
1
)
(
2
m
)
+
(
2
1
5
)
(
m
−
2
n
)
{\displaystyle {\begin{pmatrix}4\\7\\7\end{pmatrix}}m+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}n={\begin{pmatrix}1\\3\\1\end{pmatrix}}(2m)+{\begin{pmatrix}2\\1\\5\end{pmatrix}}(m-2n)}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
These sets are equal.
To prove that
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
2
m
+
4
n
=
1
t
4
m
+
8
n
=
2
t
{\displaystyle {\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\4m&+&8n&=&2t\end{array}}}
m
{\displaystyle m}
n
{\displaystyle n}
t
{\displaystyle t}
(
2
4
1
t
4
8
2
t
)
→
−
2
ρ
1
+
ρ
2
(
2
4
1
t
0
0
0
)
{\displaystyle \left({\begin{array}{*{2}{c}|c}2&4&1t\\4&8&2t\end{array}}\right){\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}2&4&1t\\0&0&0\end{array}}\right)}
m
,
n
{\displaystyle m,n}
2
m
+
4
n
=
t
{\displaystyle 2m+4n=t}
(
1
2
)
t
=
(
2
4
)
(
(
1
/
2
)
t
)
+
(
4
8
)
(
0
)
{\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}t={\begin{pmatrix}2\\4\end{pmatrix}}((1/2)t)+{\begin{pmatrix}4\\8\end{pmatrix}}(0)}
(
1
2
)
t
=
(
2
4
)
(
(
−
3
/
2
)
t
)
+
(
4
8
)
(
t
)
.
{\displaystyle {\begin{pmatrix}1\\2\end{pmatrix}}t={\begin{pmatrix}2\\4\end{pmatrix}}((-3/2)t)+{\begin{pmatrix}4\\8\end{pmatrix}}(t).}
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
S
2
⊆
S
1
{\displaystyle S_{2}\subseteq S_{1}}
1
t
=
2
m
+
4
n
2
t
=
4
m
+
8
n
{\displaystyle {\begin{array}{*{2}{rc}r}1t&=&2m+4n\\2t&=&4m+8n\end{array}}}
(
1
2
m
+
4
n
2
4
m
+
8
n
)
→
−
2
ρ
1
+
ρ
2
(
1
2
m
+
4
n
0
0
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{1}{c}|c}1&2m+4n\\2&4m+8n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{1}{c}|c}1&2m+4n\\0&0\end{array}}\right)\end{array}}}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
(
2
4
)
m
+
(
4
8
)
n
=
(
1
2
)
(
2
m
+
4
n
)
∈
S
1
.
{\displaystyle {\begin{pmatrix}2\\4\end{pmatrix}}m+{\begin{pmatrix}4\\8\end{pmatrix}}n={\begin{pmatrix}1\\2\end{pmatrix}}(2m+4n)\in S_{1}.}
Neither set is a subset of the other.
For
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
−
1
m
+
0
n
=
1
s
−
1
t
1
m
+
1
n
=
0
s
+
1
t
1
m
+
3
n
=
2
s
+
0
t
{\displaystyle {\begin{array}{*{2}{rc}r}-1m&+&0n&=&1s-1t\\1m&+&1n&=&0s+1t\\1m&+&3n&=&2s+0t\end{array}}}
m
{\displaystyle m}
n
{\displaystyle n}
t
{\displaystyle t}
s
{\displaystyle s}
(
−
1
0
1
s
−
1
t
1
1
0
s
+
1
t
1
3
2
s
+
0
t
)
→
ρ
1
+
ρ
3
ρ
1
+
ρ
2
(
−
1
0
1
s
−
1
t
0
1
1
s
+
0
t
0
3
3
s
−
1
t
)
→
−
3
ρ
2
+
ρ
3
(
−
1
0
1
s
−
1
t
0
1
1
s
+
0
t
0
3
0
s
−
1
t
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\1&1&0s+1t\\1&3&2s+0t\end{array}}\right)&{\xrightarrow[{\rho _{1}+\rho _{3}}]{\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&3s-1t\end{array}}\right)\\&{\xrightarrow[{}]{-3\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&0s-1t\end{array}}\right)\end{array}}}
m
,
n
{\displaystyle m,n}
t
=
0
{\displaystyle t=0}
(
−
1
1
0
)
{\displaystyle {\begin{pmatrix}-1\\1\\0\end{pmatrix}}}
(
−
1
1
1
)
m
+
(
0
1
3
)
n
.
{\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}n.}
S
1
{\displaystyle S_{1}}
S
2
{\displaystyle S_{2}}
S
1
≠
S
2
{\displaystyle S_{1}\neq S_{2}}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
S
2
⊆
S
1
{\displaystyle S_{2}\subseteq S_{1}}
1
s
−
1
t
=
−
1
m
+
0
n
0
s
+
1
t
=
1
m
+
1
n
2
s
+
0
t
=
1
m
+
3
n
{\displaystyle {\begin{array}{*{2}{rc}r}1s&-&1t&=&-1m+0n\\0s&+&1t&=&1m+1n\\2s&+&0t&=&1m+3n\end{array}}}
s
{\displaystyle s}
t
{\displaystyle t}
(
1
−
1
−
1
m
+
0
n
0
1
1
m
+
1
n
2
0
1
m
+
3
n
)
→
−
2
ρ
1
+
ρ
3
(
1
−
1
−
1
m
+
0
n
0
1
1
m
+
1
n
0
2
3
m
+
3
n
)
→
−
2
ρ
2
+
ρ
3
(
1
−
1
−
1
m
+
0
n
0
1
1
m
+
1
n
0
0
1
m
+
1
n
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\2&0&1m+3n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&2&3m+3n\end{array}}\right)\\&{\xrightarrow[{}]{-2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&0&1m+1n\end{array}}\right)\end{array}}}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
(
−
1
1
1
)
m
+
(
0
1
3
)
(
−
m
)
.
{\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}(-m).}
(
−
1
1
1
)
{\displaystyle {\begin{pmatrix}-1\\1\\1\end{pmatrix}}}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
These sets are equal.
First we change the parameters:
S
2
=
{
(
3
7
7
)
m
+
(
1
3
1
)
n
|
m
,
n
∈
R
}
.
{\displaystyle S_{2}=\{{\begin{pmatrix}3\\7\\7\end{pmatrix}}m+{\begin{pmatrix}1\\3\\1\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}.}
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
3
m
+
1
n
=
1
t
+
2
s
7
m
+
3
n
=
3
t
+
4
s
7
m
+
1
n
=
1
t
+
6
s
{\displaystyle {\begin{array}{*{2}{rc}r}3m&+&1n&=&1t+2s\\7m&+&3n&=&3t+4s\\7m&+&1n&=&1t+6s\end{array}}}
(
3
1
1
t
+
2
s
7
3
3
t
+
4
s
7
1
1
t
+
6
s
)
→
(
−
7
/
3
)
ρ
1
+
ρ
3
(
−
7
/
3
)
ρ
1
+
ρ
2
(
3
1
1
t
+
2
s
0
2
/
3
(
2
/
3
)
t
−
(
2
/
3
)
s
0
−
4
/
3
(
−
4
/
3
)
t
+
(
4
/
3
)
s
)
→
2
ρ
2
+
ρ
3
(
3
1
1
t
+
2
s
0
2
/
3
(
2
/
3
)
t
−
(
2
/
3
)
s
0
0
0
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\7&3&3t+4s\\7&1&1t+6s\end{array}}\right)&{\xrightarrow[{(-7/3)\rho _{1}+\rho _{3}}]{(-7/3)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&-4/3&(-4/3)t+(4/3)s\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&0&0\end{array}}\right)\end{array}}}
(
1
3
1
)
t
+
(
2
4
6
)
s
=
(
3
7
7
)
(
s
)
+
(
1
3
1
)
(
t
−
s
)
{\displaystyle {\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\4\\6\end{pmatrix}}s={\begin{pmatrix}3\\7\\7\end{pmatrix}}(s)+{\begin{pmatrix}1\\3\\1\end{pmatrix}}(t-s)}
S
1
⊆
S
2
{\displaystyle S_{1}\subseteq S_{2}}
S
2
⊆
S
1
{\displaystyle S_{2}\subseteq S_{1}}
1
t
+
2
s
=
3
m
+
1
n
3
t
+
4
s
=
7
m
+
3
n
1
t
+
6
s
=
7
m
+
1
n
{\displaystyle {\begin{array}{*{2}{rc}r}1t&+&2s&=&3m+1n\\3t&+&4s&=&7m+3n\\1t&+&6s&=&7m+1n\end{array}}}
(
1
2
3
m
+
1
n
3
4
7
m
+
3
n
1
6
7
m
+
1
n
)
→
−
ρ
1
+
ρ
3
−
3
ρ
1
+
ρ
2
(
1
2
3
m
+
1
n
0
−
2
−
2
m
0
4
4
m
)
→
2
ρ
2
+
ρ
3
(
1
2
3
m
+
1
n
0
−
2
−
2
m
0
0
0
)
{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\3&4&7m+3n\\1&6&7m+1n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&4&4m\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&0&0\end{array}}\right)\end{array}}}
(
3
7
7
)
m
+
(
1
3
1
)
n
=
(
1
3
1
)
(
m
+
n
)
+
(
2
4
6
)
(
m
)
{\displaystyle {\begin{pmatrix}3\\7\\7\end{pmatrix}}m+{\begin{pmatrix}1\\3\\1\end{pmatrix}}n={\begin{pmatrix}1\\3\\1\end{pmatrix}}(m+n)+{\begin{pmatrix}2\\4\\6\end{pmatrix}}(m)}
S
2
{\displaystyle S_{2}}
S
1
{\displaystyle S_{1}}
![{\displaystyle {\begin{array}{rcl}&{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}&{\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\&&y&&&+&w&=&5\end{array}}\\&{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}&{\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\&&&&&&2w&=&2\end{array}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08ad40a05a4e11f239161f9502583b8d8d608049)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\7&-2&3t+1s\\7&-10&1t+5s\end{array}}\right)&{\xrightarrow[{(-7/4)\rho _{1}+\rho _{3}}]{(-7/4)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&-3&-(3/4)t+(6/4)s\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}4&-4&1t+2s\\0&5&(5/4)t-(10/4)s\\0&0&0\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18e6345a99a0dda5217f3a50004014cc6c6e9f07)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\3&1&7m-2n\\1&5&7m-10n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&3&3m-6n\end{array}}\right)\\&{\xrightarrow[{}]{(3/5)\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&4m-4n\\0&-5&-5m+10n\\0&0&0\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ce7bc4f6039965e319643f7f333b0484d007e63)
![{\displaystyle \left({\begin{array}{*{2}{c}|c}2&4&1t\\4&8&2t\end{array}}\right){\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}2&4&1t\\0&0&0\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45166a9712ae2874df071cc021ceb2f97644b8d4)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{1}{c}|c}1&2m+4n\\2&4m+8n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{1}{c}|c}1&2m+4n\\0&0\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12e6fbbbb8d73738a326bdd481c22ec2ce9574c8)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\1&1&0s+1t\\1&3&2s+0t\end{array}}\right)&{\xrightarrow[{\rho _{1}+\rho _{3}}]{\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&3s-1t\end{array}}\right)\\&{\xrightarrow[{}]{-3\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}-1&0&1s-1t\\0&1&1s+0t\\0&3&0s-1t\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fb46ca62de5c1148a5800e53387219825aea67e)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\2&0&1m+3n\end{array}}\right)&{\xrightarrow[{}]{-2\rho _{1}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&2&3m+3n\end{array}}\right)\\&{\xrightarrow[{}]{-2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&-1&-1m+0n\\0&1&1m+1n\\0&0&1m+1n\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59fbce7a4a56260621facf17aec2463054026aee)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\7&3&3t+4s\\7&1&1t+6s\end{array}}\right)&{\xrightarrow[{(-7/3)\rho _{1}+\rho _{3}}]{(-7/3)\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&-4/3&(-4/3)t+(4/3)s\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}3&1&1t+2s\\0&2/3&(2/3)t-(2/3)s\\0&0&0\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ffde0a4dbfa0a4a867f9c09b561a4f71fba90f2)
![{\displaystyle {\begin{array}{rcl}\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\3&4&7m+3n\\1&6&7m+1n\end{array}}\right)&{\xrightarrow[{-\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&4&4m\end{array}}\right)\\&{\xrightarrow[{}]{2\rho _{2}+\rho _{3}}}&\left({\begin{array}{*{2}{c}|c}1&2&3m+1n\\0&-2&-2m\\0&0&0\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f4d43cd9953135aad7368b1b651b928e1e29fb9)
