This exercise is recommended for all readers. This exercise is recommended for all readers. This exercise is recommended for all readers.
Problem 3
Find each solution set by using Gauss-Jordan reduction,
then reading off the parametrization.
2
x
+
y
−
z
=
1
4
x
−
y
=
3
{\displaystyle {\begin{array}{*{3}{rc}r}2x&+&y&-&z&=&1\\4x&-&y&&&=&3\end{array}}}
x
−
z
=
1
y
+
2
z
−
w
=
3
x
+
2
y
+
3
z
−
w
=
7
{\displaystyle {\begin{array}{*{4}{rc}r}x&&&-&z&&&=&1\\&&y&+&2z&-&w&=&3\\x&+&2y&+&3z&-&w&=&7\end{array}}}
x
−
y
+
z
=
0
y
+
w
=
0
3
x
−
2
y
+
3
z
+
w
=
0
−
y
−
w
=
0
{\displaystyle {\begin{array}{*{4}{rc}r}x&-&y&+&z&&&=&0\\&&y&&&+&w&=&0\\3x&-&2y&+&3z&+&w&=&0\\&&-y&&&-&w&=&0\end{array}}}
a
+
2
b
+
3
c
+
d
−
e
=
1
3
a
−
b
+
c
+
d
+
e
=
3
{\displaystyle {\begin{array}{*{5}{rc}r}a&+&2b&+&3c&+&d&-&e&=&1\\3a&-&b&+&c&+&d&+&e&=&3\end{array}}}
Answer
For the "Gauss" halves, see the answers to
Problem I.2.5 .
The "Jordan" half goes this way.
→
−
(
1
/
3
)
ρ
2
(
1
/
2
)
ρ
1
(
1
1
/
2
−
1
/
2
1
/
2
0
1
−
2
/
3
−
1
/
3
)
→
−
(
1
/
2
)
ρ
2
+
ρ
1
(
1
0
−
1
/
6
2
/
3
0
1
−
2
/
3
−
1
/
3
)
{\displaystyle {\xrightarrow[{-(1/3)\rho _{2}}]{(1/2)\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&1/2&-1/2&1/2\\0&1&-2/3&-1/3\end{array}}\right){\xrightarrow[{}]{-(1/2)\rho _{2}+\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&0&-1/6&2/3\\0&1&-2/3&-1/3\end{array}}\right)}
{
(
2
/
3
−
1
/
3
0
)
+
(
1
/
6
2
/
3
1
)
z
|
z
∈
R
}
{\displaystyle \{{\begin{pmatrix}2/3\\-1/3\\0\end{pmatrix}}+{\begin{pmatrix}1/6\\2/3\\1\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}}
The second half is
→
ρ
3
+
ρ
2
(
1
0
−
1
0
1
0
1
2
0
3
0
0
0
1
0
)
{\displaystyle {\xrightarrow[{}]{\rho _{3}+\rho _{2}}}\left({\begin{array}{*{4}{c}|c}1&0&-1&0&1\\0&1&2&0&3\\0&0&0&1&0\end{array}}\right)}
{
(
1
3
0
0
)
+
(
1
−
2
1
0
)
z
|
z
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\3\\0\\0\end{pmatrix}}+{\begin{pmatrix}1\\-2\\1\\0\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}}
This Jordan half
→
ρ
2
+
ρ
1
(
1
0
1
1
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
)
{\displaystyle {\xrightarrow[{}]{\rho _{2}+\rho _{1}}}\left({\begin{array}{*{4}{c}|c}1&0&1&1&0\\0&1&0&1&0\\0&0&0&0&0\\0&0&0&0&0\end{array}}\right)}
{
(
0
0
0
0
)
+
(
−
1
0
1
0
)
z
+
(
−
1
−
1
0
1
)
w
|
z
,
w
∈
R
}
{\displaystyle \{{\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}z+{\begin{pmatrix}-1\\-1\\0\\1\end{pmatrix}}w\,{\big |}\,z,w\in \mathbb {R} \}}
The "Jordan" half
→
−
(
1
/
7
)
ρ
2
(
1
2
3
1
−
1
1
0
1
8
/
7
2
/
7
−
4
/
7
0
)
→
−
2
ρ
2
+
ρ
1
(
1
0
5
/
7
3
/
7
1
/
7
1
0
1
8
/
7
2
/
7
−
4
/
7
0
)
{\displaystyle {\xrightarrow[{}]{-(1/7)\rho _{2}}}\left({\begin{array}{*{5}{c}|c}1&2&3&1&-1&1\\0&1&8/7&2/7&-4/7&0\end{array}}\right){\xrightarrow[{}]{-2\rho _{2}+\rho _{1}}}\left({\begin{array}{*{5}{c}|c}1&0&5/7&3/7&1/7&1\\0&1&8/7&2/7&-4/7&0\end{array}}\right)}
{
(
1
0
0
0
0
)
+
(
−
5
/
7
−
8
/
7
1
0
0
)
c
+
(
−
3
/
7
−
2
/
7
0
1
0
)
d
+
(
−
1
/
7
4
/
7
0
0
1
)
e
|
c
,
d
,
e
∈
R
}
{\displaystyle \{{\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix}}+{\begin{pmatrix}-5/7\\-8/7\\1\\0\\0\end{pmatrix}}c+{\begin{pmatrix}-3/7\\-2/7\\0\\1\\0\end{pmatrix}}d+{\begin{pmatrix}-1/7\\4/7\\0\\0\\1\end{pmatrix}}e\,{\big |}\,c,d,e\in \mathbb {R} \}}
Problem 4
Give two distinct echelon form versions of this matrix.
(
2
1
1
3
6
4
1
2
1
5
1
5
)
{\displaystyle {\begin{pmatrix}2&1&1&3\\6&4&1&2\\1&5&1&5\end{pmatrix}}}
Answer
Routine Gauss' method gives one:
→
−
(
1
/
2
)
ρ
1
+
ρ
3
−
3
ρ
1
+
ρ
2
(
2
1
1
3
0
1
−
2
−
7
0
9
/
2
1
/
2
7
/
2
)
→
−
(
9
/
2
)
ρ
2
+
ρ
3
(
2
1
1
3
0
1
−
2
−
7
0
0
19
/
2
35
)
{\displaystyle {\xrightarrow[{-(1/2)\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}{\begin{pmatrix}2&1&1&3\\0&1&-2&-7\\0&9/2&1/2&7/2\end{pmatrix}}{\xrightarrow[{}]{-(9/2)\rho _{2}+\rho _{3}}}{\begin{pmatrix}2&1&1&3\\0&1&-2&-7\\0&0&19/2&35\end{pmatrix}}}
and any cosmetic change, like multiplying the bottom row by
2
{\displaystyle 2}
(
2
1
1
3
0
1
−
2
−
7
0
0
19
70
)
{\displaystyle {\begin{pmatrix}2&1&1&3\\0&1&-2&-7\\0&0&19&70\end{pmatrix}}}
gives another.
This exercise is recommended for all readers.
Problem 5
List the reduced echelon forms possible for each size.
2
×
2
{\displaystyle 2\!\times \!2}
2
×
3
{\displaystyle 2\!\times \!3}
3
×
2
{\displaystyle 3\!\times \!2}
3
×
3
{\displaystyle 3\!\times \!3}
Answer
In the cases listed below, we take
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
a
=
0
{\displaystyle a=0}
b
=
0
{\displaystyle b=0}
(
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0\\0&0\end{pmatrix}}}
(
1
a
0
0
)
{\displaystyle {\begin{pmatrix}1&a\\0&0\end{pmatrix}}}
(
0
1
0
0
)
{\displaystyle {\begin{pmatrix}0&1\\0&0\end{pmatrix}}}
(
1
0
0
1
)
{\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}}
(
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0&0\\0&0&0\end{pmatrix}}}
(
1
a
b
0
0
0
)
{\displaystyle {\begin{pmatrix}1&a&b\\0&0&0\end{pmatrix}}}
(
0
1
a
0
0
0
)
{\displaystyle {\begin{pmatrix}0&1&a\\0&0&0\end{pmatrix}}}
(
0
0
1
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0&1\\0&0&0\end{pmatrix}}}
(
1
0
a
0
1
b
)
{\displaystyle {\begin{pmatrix}1&0&a\\0&1&b\end{pmatrix}}}
(
1
a
0
0
0
1
)
{\displaystyle {\begin{pmatrix}1&a&0\\0&0&1\end{pmatrix}}}
(
0
1
0
0
0
1
)
{\displaystyle {\begin{pmatrix}0&1&0\\0&0&1\end{pmatrix}}}
(
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0\\0&0\\0&0\end{pmatrix}}}
(
1
a
0
0
0
0
)
{\displaystyle {\begin{pmatrix}1&a\\0&0\\0&0\end{pmatrix}}}
(
0
1
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&1\\0&0\\0&0\end{pmatrix}}}
(
1
0
0
1
0
0
)
{\displaystyle {\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix}}}
(
0
0
0
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}}}
(
1
a
b
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}1&a&b\\0&0&0\\0&0&0\end{pmatrix}}}
(
0
1
a
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&1&a\\0&0&0\\0&0&0\end{pmatrix}}}
(
0
0
1
0
0
0
0
0
0
)
{\displaystyle {\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}}}
(
1
0
a
0
1
b
0
0
0
)
{\displaystyle {\begin{pmatrix}1&0&a\\0&1&b\\0&0&0\end{pmatrix}}}
(
1
a
0
0
0
1
0
0
0
)
{\displaystyle {\begin{pmatrix}1&a&0\\0&0&1\\0&0&0\end{pmatrix}}}
(
0
1
0
0
0
1
0
0
0
)
{\displaystyle {\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}}
(
1
0
0
0
1
0
0
0
1
)
{\displaystyle {\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}
This exercise is recommended for all readers.
Problem 6
What results from applying Gauss-Jordan reduction to a
nonsingular matrix?
Answer
A nonsingular homogeneous linear system has a unique solution.
So a nonsingular matrix must reduce to a (square)
matrix that is all
0
{\displaystyle 0}
1
{\displaystyle 1}
(
1
0
0
1
)
,
or
(
1
0
0
0
1
0
0
0
1
)
,
etc.
{\displaystyle {\begin{pmatrix}1&0\\0&1\\\end{pmatrix}},\quad {\text{or}}\quad {\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}},\quad {\text{etc.}}}
Problem 7
The proof of Lemma 4 contains a reference to the
i
≠
j
{\displaystyle i\neq j}
The definition of row operations has an
i
≠
j
{\displaystyle i\neq j}
ρ
i
↔
ρ
j
{\displaystyle \rho _{i}\leftrightarrow \rho _{j}}
A
→
ρ
i
↔
ρ
j
→
ρ
i
↔
ρ
j
A
{\displaystyle A{\xrightarrow[{}]{\rho _{i}\leftrightarrow \rho _{j}}}\;{\xrightarrow[{}]{\rho _{i}\leftrightarrow \rho _{j}}}A}
Write down a
2
×
2
{\displaystyle 2\!\times \!2}
−
1
⋅
ρ
1
+
ρ
1
{\displaystyle -1\cdot \rho _{1}+\rho _{1}}
1
⋅
ρ
1
+
ρ
1
{\displaystyle 1\cdot \rho _{1}+\rho _{1}}
Expand the proof of that lemma to make explicit exactly where
the
i
≠
j
{\displaystyle i\neq j}
Answer
The
ρ
i
↔
ρ
i
{\displaystyle \rho _{i}\leftrightarrow \rho _{i}}
A
{\displaystyle A}
For instance,
(
1
2
3
4
)
→
−
ρ
1
+
ρ
1
(
0
0
3
4
)
→
ρ
1
+
ρ
1
(
0
0
3
4
)
{\displaystyle {\begin{pmatrix}1&2\\3&4\end{pmatrix}}{\xrightarrow[{}]{-\rho _{1}+\rho _{1}}}{\begin{pmatrix}0&0\\3&4\end{pmatrix}}{\xrightarrow[{}]{\rho _{1}+\rho _{1}}}{\begin{pmatrix}0&0\\3&4\end{pmatrix}}}
If
i
≠
j
{\displaystyle i\neq j}
(
⋮
a
i
,
1
⋯
a
i
,
n
⋮
a
j
,
1
⋯
a
j
,
n
⋮
)
→
k
ρ
i
+
ρ
j
(
⋮
a
i
,
1
⋯
a
i
,
n
⋮
k
a
i
,
1
+
a
j
,
1
⋯
k
a
i
,
n
+
a
j
,
n
⋮
)
→
−
k
ρ
i
+
ρ
j
(
⋮
a
i
,
1
⋯
a
i
,
n
⋮
−
k
a
i
,
1
+
k
a
i
,
1
+
a
j
,
1
⋯
−
k
a
i
,
n
+
k
a
i
,
n
+
a
j
,
n
⋮
)
{\displaystyle {\begin{array}{rcl}{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\a_{j,1}&\cdots &a_{j,n}\\\vdots \end{pmatrix}}&{\xrightarrow[{}]{k\rho _{i}+\rho _{j}}}&{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\ka_{i,1}+a_{j,1}&\cdots &ka_{i,n}+a_{j,n}\\\vdots \end{pmatrix}}\\&{\xrightarrow[{}]{-k\rho _{i}+\rho _{j}}}&{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\-ka_{i,1}+ka_{i,1}+a_{j,1}&\cdots &-ka_{i,n}+ka_{i,n}+a_{j,n}\\\vdots \end{pmatrix}}\end{array}}}
A
{\displaystyle A}
i
=
j
{\displaystyle i=j}
−
k
(
k
a
i
,
j
+
a
i
,
j
)
+
k
a
i
,
j
+
a
i
,
j
{\displaystyle -k(ka_{i,j}+a_{i,j})+ka_{i,j}+a_{i,j}}
![{\displaystyle \left({\begin{array}{*{2}{c}|c}1&1&2\\1&-1&0\end{array}}\right){\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}1&1&2\\0&-2&-2\end{array}}\right){\xrightarrow[{}]{-(1/2)\rho _{2}}}\left({\begin{array}{*{2}{c}|c}1&1&2\\0&1&1\end{array}}\right){\xrightarrow[{}]{-\rho _{2}+\rho _{1}}}\left({\begin{array}{*{2}{c}|c}1&0&1\\0&1&1\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f6bb0ca1fec64b41a9d24ec6efa2c947b9424aa)
![{\displaystyle \left({\begin{array}{*{3}{c}|c}1&0&-1&4\\2&2&0&1\end{array}}\right){\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{3}{c}|c}1&0&-1&4\\0&2&2&-7\end{array}}\right){\xrightarrow[{}]{(1/2)\rho _{2}}}\left({\begin{array}{*{3}{c}|c}1&0&-1&4\\0&1&1&-7/2\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ae33612bd33db97ba4cbaf0d919880ba2291610)
![{\displaystyle \left({\begin{array}{*{2}{c}|c}3&-2&1\\6&1&1/2\end{array}}\right){\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}\left({\begin{array}{*{2}{c}|c}3&-2&1\\0&5&-3/2\end{array}}\right){\xrightarrow[{(1/5)\rho _{2}}]{(1/3)\rho _{1}}}\left({\begin{array}{*{2}{c}|c}1&-2/3&1/3\\0&1&-3/10\end{array}}\right){\xrightarrow[{}]{(2/3)\rho _{2}+\rho _{1}}}\left({\begin{array}{*{2}{c}|c}1&0&2/15\\0&1&-3/10\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82bb7ff0ff3710143e98002971d25af31cae0bff)
![{\displaystyle \left({\begin{array}{*{3}{c}|c}2&-1&0&-1\\1&3&-1&5\\0&1&2&5\end{array}}\right){\xrightarrow[{}]{-(1/2)\rho _{1}+\rho _{2}}}\left({\begin{array}{*{3}{c}|c}2&-1&0&-1\\0&7/2&-1&11/2\\0&1&2&5\end{array}}\right){\xrightarrow[{}]{\rho _{2}\leftrightarrow \rho _{3}}}\left({\begin{array}{*{3}{c}|c}2&-1&0&-1\\0&1&2&5\\0&7/2&-1&11/2\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c3218fa5597b28d40168edde342c65173287f86)
![{\displaystyle {\begin{array}{rl}&{\xrightarrow[{}]{-(7/2)\rho _{2}+\rho _{3}}}\left({\begin{array}{*{3}{c}|c}2&-1&0&-1\\0&1&2&5\\0&0&-8&-12\end{array}}\right){\xrightarrow[{-(1/8)\rho _{2}}]{(1/2)\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&-1/2&0&-1/2\\0&1&2&5\\0&0&1&3/2\end{array}}\right)\\&{\xrightarrow[{}]{-2\rho _{3}+\rho _{2}}}\left({\begin{array}{*{3}{c}|c}1&-1/2&0&-1/2\\0&1&0&2\\0&0&1&3/2\end{array}}\right){\xrightarrow[{}]{(1/2)\rho _{2}+\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&0&0&1/2\\0&1&0&2\\0&0&1&3/2\end{array}}\right)\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/388fe494ac0eab290912f5c4aa6fe15a69c298fd)
![{\displaystyle {\xrightarrow[{}]{-(1/2)\rho _{1}+\rho _{2}}}{\begin{pmatrix}2&1\\0&5/2\end{pmatrix}}{\xrightarrow[{(2/5)\rho _{2}}]{(1/2)\rho _{1}}}{\begin{pmatrix}1&1/2\\0&1\end{pmatrix}}{\xrightarrow[{}]{-(1/2)\rho _{2}+\rho _{1}}}{\begin{pmatrix}1&0\\0&1\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83168e90412703d467bdc00007c997518bda68f2)
![{\displaystyle {\xrightarrow[{\rho _{1}+\rho _{3}}]{-2\rho _{1}+\rho _{2}}}{\begin{pmatrix}1&3&1\\0&-6&2\\0&0&-2\end{pmatrix}}{\xrightarrow[{-(1/2)\rho _{3}}]{-(1/6)\rho _{2}}}{\begin{pmatrix}1&3&1\\0&1&-1/3\\0&0&1\end{pmatrix}}{\xrightarrow[{-\rho _{3}+\rho _{1}}]{(1/3)\rho _{3}+\rho _{2}}}{\begin{pmatrix}1&3&0\\0&1&0\\0&0&1\end{pmatrix}}{\xrightarrow[{}]{-3\rho _{2}+\rho _{1}}}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2eea8d9b9a24c966312eb3d1a4a57bb0cb738dd9)
![{\displaystyle {\xrightarrow[{-3\rho _{1}+\rho _{3}}]{-\rho _{1}+\rho _{2}}}{\begin{pmatrix}1&0&3&1&2\\0&4&-1&0&3\\0&4&-1&-2&-4\end{pmatrix}}{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}{\begin{pmatrix}1&0&3&1&2\\0&4&-1&0&3\\0&0&0&-2&-7\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a73064c6a4e4f0ba227c6bfdb7cd603bd0135fe)
![{\displaystyle {\xrightarrow[{-(1/2)\rho _{3}}]{(1/4)\rho _{2}}}{\begin{pmatrix}1&0&3&1&2\\0&1&-1/4&0&3/4\\0&0&0&1&7/2\end{pmatrix}}{\xrightarrow[{}]{-\rho _{3}+\rho _{1}}}{\begin{pmatrix}1&0&3&0&-3/2\\0&1&-1/4&0&3/4\\0&0&0&1&7/2\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac319a7b00d81877e8f70904e97c1328d2fc542f)
![{\displaystyle {\xrightarrow[{}]{\rho _{1}\leftrightarrow \rho _{3}}}{\begin{pmatrix}1&5&1&5\\0&0&5&6\\0&1&3&2\end{pmatrix}}{\xrightarrow[{}]{\rho _{2}\leftrightarrow \rho _{3}}}{\begin{pmatrix}1&5&1&5\\0&1&3&2\\0&0&5&6\end{pmatrix}}{\xrightarrow[{}]{(1/5)\rho _{3}}}{\begin{pmatrix}1&5&1&5\\0&1&3&2\\0&0&1&6/5\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4c5f9336d7b20c475ef7e34556fbabf05a055ed)
![{\displaystyle {\xrightarrow[{-\rho _{3}+\rho _{1}}]{-3\rho _{3}+\rho _{2}}}{\begin{pmatrix}1&5&0&19/5\\0&1&0&-8/5\\0&0&1&6/5\end{pmatrix}}{\xrightarrow[{}]{-5\rho _{2}+\rho _{1}}}{\begin{pmatrix}1&0&0&59/5\\0&1&0&-8/5\\0&0&1&6/5\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16674ad873690eb5122f4bf2d7ae2371a2e50921)
![{\displaystyle {\xrightarrow[{-(1/3)\rho _{2}}]{(1/2)\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&1/2&-1/2&1/2\\0&1&-2/3&-1/3\end{array}}\right){\xrightarrow[{}]{-(1/2)\rho _{2}+\rho _{1}}}\left({\begin{array}{*{3}{c}|c}1&0&-1/6&2/3\\0&1&-2/3&-1/3\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7acd26ad9540757c3d604ca67b2db268cd270c9f)
![{\displaystyle {\xrightarrow[{}]{\rho _{3}+\rho _{2}}}\left({\begin{array}{*{4}{c}|c}1&0&-1&0&1\\0&1&2&0&3\\0&0&0&1&0\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d3b57d5c2305b62ecda0cda5024f9709ba93483)
![{\displaystyle {\xrightarrow[{}]{\rho _{2}+\rho _{1}}}\left({\begin{array}{*{4}{c}|c}1&0&1&1&0\\0&1&0&1&0\\0&0&0&0&0\\0&0&0&0&0\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9283bc5b4ccc9f6fa973ec2f18ab663fe2e46f0e)
![{\displaystyle {\xrightarrow[{}]{-(1/7)\rho _{2}}}\left({\begin{array}{*{5}{c}|c}1&2&3&1&-1&1\\0&1&8/7&2/7&-4/7&0\end{array}}\right){\xrightarrow[{}]{-2\rho _{2}+\rho _{1}}}\left({\begin{array}{*{5}{c}|c}1&0&5/7&3/7&1/7&1\\0&1&8/7&2/7&-4/7&0\end{array}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7552bfc67375a5494d65986195eccb79e39a222b)
![{\displaystyle {\xrightarrow[{-(1/2)\rho _{1}+\rho _{3}}]{-3\rho _{1}+\rho _{2}}}{\begin{pmatrix}2&1&1&3\\0&1&-2&-7\\0&9/2&1/2&7/2\end{pmatrix}}{\xrightarrow[{}]{-(9/2)\rho _{2}+\rho _{3}}}{\begin{pmatrix}2&1&1&3\\0&1&-2&-7\\0&0&19/2&35\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c06334a6dcb9481448cd08e320b4f5ab76c1a01)
![{\displaystyle A{\xrightarrow[{}]{\rho _{i}\leftrightarrow \rho _{j}}}\;{\xrightarrow[{}]{\rho _{i}\leftrightarrow \rho _{j}}}A}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d75ef0de0af637fe8d2a4e94862463e1a9c9947d)
![{\displaystyle {\begin{pmatrix}1&2\\3&4\end{pmatrix}}{\xrightarrow[{}]{-\rho _{1}+\rho _{1}}}{\begin{pmatrix}0&0\\3&4\end{pmatrix}}{\xrightarrow[{}]{\rho _{1}+\rho _{1}}}{\begin{pmatrix}0&0\\3&4\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c87efb1de70f7ba0690a407440d674a388f7d7e)
![{\displaystyle {\begin{array}{rcl}{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\a_{j,1}&\cdots &a_{j,n}\\\vdots \end{pmatrix}}&{\xrightarrow[{}]{k\rho _{i}+\rho _{j}}}&{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\ka_{i,1}+a_{j,1}&\cdots &ka_{i,n}+a_{j,n}\\\vdots \end{pmatrix}}\\&{\xrightarrow[{}]{-k\rho _{i}+\rho _{j}}}&{\begin{pmatrix}\vdots \\a_{i,1}&\cdots &a_{i,n}\\\vdots \\-ka_{i,1}+ka_{i,1}+a_{j,1}&\cdots &-ka_{i,n}+ka_{i,n}+a_{j,n}\\\vdots \end{pmatrix}}\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ef8d6e7687a98315aab9a8b0991c6e943e3aebe)
