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Razširjena matrika je v linearni algebri matrika , ki se dobi z dodajanjem stolpcev dani matriki. To naredimo zaradi izvajanja istih elementarnih vrstičnih operacij na obeh matrikah.
Za dani matriki
A
{\displaystyle A\,}
B
{\displaystyle B\,}
A
=
[
1
3
2
2
0
1
5
2
2
]
,
B
=
[
4
3
1
]
.
{\displaystyle A={\begin{bmatrix}1&3&2\\2&0&1\\5&2&2\end{bmatrix}},\quad B={\begin{bmatrix}4\\3\\1\end{bmatrix}}.}
se razširjena matrika piše kot
(
A
|
B
)
=
[
1
3
2
4
2
0
1
3
5
2
2
1
]
.
{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}
Takšen način pisanja je zelo uporaben pri reševanju sistema linearnih enačb in pri določanju obratnih matrik .
Imamo kvadratno matriko z razsežnostjo
2
×
2
{\displaystyle 2\times 2\,}
C
=
[
1
3
−
5
0
]
.
{\displaystyle C={\begin{bmatrix}1&3\\-5&0\end{bmatrix}}.}
Za določanje obratne matrike moramo narediti razširjeno matriko
C
|
I
{\displaystyle C|I\,}
I
{\displaystyle I\,}
enotska matrika z razsežnostjo
2
×
2
{\displaystyle 2\times 2\,}
C
−
1
{\displaystyle C^{-1}\,}
C
{\displaystyle C\,}
(
C
|
I
)
=
[
1
3
1
0
−
5
0
0
1
]
{\displaystyle (C|I)=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}
(
I
|
C
−
1
)
=
[
1
0
0
−
1
5
0
1
1
3
1
15
]
{\displaystyle (I|C^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right]}
Dane imamo tri linearne enačbe
x
1
+
2
x
2
+
3
x
3
=
0
3
x
1
+
4
x
2
+
7
x
3
=
2
6
x
1
+
5
x
2
+
9
x
3
=
11
{\displaystyle {\begin{aligned}x_{1}+2x_{2}+3x_{3}&=0\\3x_{1}+4x_{2}+7x_{3}&=2\\6x_{1}+5x_{2}+9x_{3}&=11\end{aligned}}}
To pa odgovarja naslednjim matrikam
A
=
[
1
2
3
3
4
7
6
5
9
]
,
B
=
[
0
2
11
]
.
{\displaystyle A={\begin{bmatrix}1&2&3\\3&4&7\\6&5&9\end{bmatrix}},\quad B={\begin{bmatrix}0\\2\\11\end{bmatrix}}.}
(
A
|
B
)
=
[
1
2
3
0
3
4
7
2
6
5
9
11
]
,
{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right],}
ali
(
A
|
B
)
=
[
1
0
0
4
0
1
0
1
0
0
1
−
2
]
.
{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right].}
To pomeni, da so rešitve zgornjega sistema linearnih enačb
x
1
=
4
{\displaystyle x_{1}=4\,}
x
2
=
1
{\displaystyle x_{2}=1\,}
x
3
=
−
2
{\displaystyle x_{3}=-2\,}
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f14272674b4b17491481e7ae9bcfcbc168ccb58)
![{\displaystyle (C|I)=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f5ad0f062318891216bc045d12b5fa019c00a8d)
![{\displaystyle (I|C^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59be87e78725b558b9c1baf01edc097e9cb7ed7f)
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d824a756a96bc77970c21efb74a2f82de1bbe9a)
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3c8beb24663ffb74a5d6fcb7d111c218e51f3cb)
