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40 eilutė:
40 eilutė:
Sudauginę ''A'' matricą su jos atvirkštine matrica <math>A^{-1}</math>, gauname vienetinę matricą:
Sudauginę ''A'' matricą su jos atvirkštine matrica <math>A^{-1}</math>, gauname vienetinę matricą:
:<math>E=A A^{-1} = \begin{bmatrix} 3 & -1 & 0 \\ -2 & 1 & 1 \\ 2 & -1 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & \frac{4}{5} & -\frac{1}{5} \\ 2 & \frac{12}{5} & -\frac{3}{5} \\ 0 & \frac{1}{5} & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 3\cdot 1+(-2)\cdot 2+0\cdot 0 & 3\cdot\frac{4}{5}+(-1)\cdot\frac{12}{5}+0\cdot \frac{1}{5} & 3\cdot(-\frac{1}{5})+(-1)\cdot(-\frac{3}{5})+0\cdot \frac{1}{5}
:<math>E=A A^{-1} = \begin{bmatrix} 3 & -1 & 0 \\ -2 & 1 & 1 \\ 2 & -1 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & \frac{4}{5} & -\frac{1}{5} \\ 2 & \frac{12}{5} & -\frac{3}{5} \\ 0 & \frac{1}{5} & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 3\cdot 1+(-1 )\cdot 2+0\cdot 0 & 3\cdot\frac{4}{5}+(-1)\cdot\frac{12}{5}+0\cdot \frac{1}{5} & 3\cdot(-\frac{1}{5})+(-1)\cdot(-\frac{3}{5})+0\cdot \frac{1}{5}
\\ -2\cdot 1+1\cdot 2+1\cdot 0 & -2\cdot \frac{4}{5}+1\cdot \frac{12}{5} +1\cdot \frac{1}{5} & -2\cdot(-\frac{1}{5}) +1\cdot(-\frac{3}{5}) + 1\cdot\frac{1}{5}
\\ -2\cdot 1+1\cdot 2+1\cdot 0 & -2\cdot \frac{4}{5}+1\cdot \frac{12}{5} +1\cdot \frac{1}{5} & -2\cdot(-\frac{1}{5}) +1\cdot(-\frac{3}{5}) + 1\cdot\frac{1}{5}
\\ 2\cdot 1 + (-1)\cdot 2+ 4\cdot 0 & 2\cdot\frac{4}{5}+ (-1)\cdot\frac{12}{5}+ 4\cdot\frac{1}{5} & 2\cdot(-\frac{1}{5}) + (-1)\cdot(-\frac{3}{5})+ 4\cdot\frac{1}{5} \end{bmatrix}=</math>
\\ 2\cdot 1 + (-1)\cdot 2+ 4\cdot 0 & 2\cdot\frac{4}{5}+ (-1)\cdot\frac{12}{5}+ 4\cdot\frac{1}{5} & 2\cdot(-\frac{1}{5}) + (-1)\cdot(-\frac{3}{5})+ 4\cdot\frac{1}{5} \end{bmatrix}=</math>
:<math>=\begin{bmatrix} 3-4+0 & \frac{12}{5}-\frac{12}{5}+0 & -\frac{3}{5}+\frac{3}{5}+0
:<math>=\begin{bmatrix} 3-2 +0 & \frac{12}{5}-\frac{12}{5}+0 & -\frac{3}{5}+\frac{3}{5}+0
\\ -2+ 2+ 0 & -\frac{8}{5}+\frac{12}{5} + \frac{1}{5} & \frac{2}{5} -\frac{3}{5} + \frac{1}{5}
\\ -2+ 2+ 0 & -\frac{8}{5}+\frac{12}{5} + \frac{1}{5} & \frac{2}{5} -\frac{3}{5} + \frac{1}{5}
\\ 2 - 2+ 0 & \frac{8}{5}- \frac{12}{5}+ \frac{4}{5} & -\frac{2}{5} + \frac{3}{5}+ \frac{4}{5} \end{bmatrix}=</math>
\\ 2 - 2+ 0 & \frac{8}{5}- \frac{12}{5}+ \frac{4}{5} & -\frac{2}{5} + \frac{3}{5}+ \frac{4}{5} \end{bmatrix}
⚫
=\begin{bmatrix} 1 & 0 & 0
⚫
:<math>=\begin{bmatrix}
-1 & 0 & 0
\\ 0 & 1 & 0
\\ 0 & 1 & 0
\\ 0 & 0 & 1 \end{bmatrix},</math>
\\ 0 & 0 & 1 \end{bmatrix},</math>
17:54, 6 gegužės 2020 versija
Atvirkštinė matrica
A
−
1
{\displaystyle A^{-1}}
yra tokia matricos A matrica, kad
A
A
−
1
=
A
−
1
=
E
,
{\displaystyle AA^{-1}=A^{-1}=E,}
Čia E yra vienetinė matrica.
A
−
1
=
1
|
A
|
⋅
[
A
11
A
21
A
31
A
12
A
22
A
32
A
13
A
23
A
33
]
,
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}},}
Čia |A| yra matricos A determinantas:
d
=
d
e
t
A
=
|
A
|
=
|
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
|
.
{\displaystyle d=detA=|A|={\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}}.}
A
11
=
(
−
1
)
1
+
1
(
a
22
a
33
−
a
23
a
32
)
,
{\displaystyle A_{11}=(-1)^{1+1}(a_{22}a_{33}-a_{23}a_{32}),}
A
21
=
(
−
1
)
2
+
1
(
a
12
a
33
−
a
13
a
32
)
,
{\displaystyle A_{21}=(-1)^{2+1}(a_{12}a_{33}-a_{13}a_{32}),}
A
31
=
(
−
1
)
3
+
1
(
a
12
a
23
−
a
13
a
22
)
,
{\displaystyle A_{31}=(-1)^{3+1}(a_{12}a_{23}-a_{13}a_{22}),}
A
12
=
(
−
1
)
1
+
2
(
a
21
a
33
−
a
23
a
31
)
,
{\displaystyle A_{12}=(-1)^{1+2}(a_{21}a_{33}-a_{23}a_{31}),}
A
22
=
(
−
1
)
2
+
2
(
a
11
a
33
−
a
13
a
31
)
,
{\displaystyle A_{22}=(-1)^{2+2}(a_{11}a_{33}-a_{13}a_{31}),}
A
32
=
(
−
1
)
3
+
2
(
a
11
a
23
−
a
13
a
21
)
,
{\displaystyle A_{32}=(-1)^{3+2}(a_{11}a_{23}-a_{13}a_{21}),}
A
13
=
(
−
1
)
1
+
3
(
a
21
a
32
−
a
22
a
31
)
,
{\displaystyle A_{13}=(-1)^{1+3}(a_{21}a_{32}-a_{22}a_{31}),}
A
23
=
(
−
1
)
2
+
3
(
a
11
a
32
−
a
12
a
31
)
,
{\displaystyle A_{23}=(-1)^{2+3}(a_{11}a_{32}-a_{12}a_{31}),}
A
33
=
(
−
1
)
3
+
3
(
a
11
a
22
−
a
12
a
21
)
.
{\displaystyle A_{33}=(-1)^{3+3}(a_{11}a_{22}-a_{12}a_{21}).}
Matricos A adjunktas
A
i
j
{\displaystyle A_{ij}}
(čia i simbolizuoja adjunkto eilutę, o j simbolizuoja adjunkto stulpelį) į atvirkštinę matricą dedamas tokiu budu, kad j reiškia eilutę atvirkštinėje matricoje, o i reiškia stulpelį.
A
=
[
3
−
1
0
−
2
1
1
2
−
1
4
]
{\displaystyle A={\begin{bmatrix}3&-1&0\\-2&1&1\\2&-1&4\end{bmatrix}}}
atvirkštinę matricą.
Pirmiausia rasime matricos A determinantą.
d
=
|
A
|
=
|
3
−
1
0
−
2
1
1
2
−
1
4
|
=
3
⋅
1
⋅
4
+
(
−
1
)
⋅
1
⋅
2
+
(
−
2
)
⋅
(
−
1
)
⋅
0
−
0
⋅
1
⋅
2
−
(
−
1
)
⋅
(
−
2
)
⋅
4
−
3
⋅
1
⋅
(
−
1
)
=
{\displaystyle d=|A|={\begin{vmatrix}3&-1&0\\-2&1&1\\2&-1&4\end{vmatrix}}=3\cdot 1\cdot 4+(-1)\cdot 1\cdot 2+(-2)\cdot (-1)\cdot 0-0\cdot 1\cdot 2-(-1)\cdot (-2)\cdot 4-3\cdot 1\cdot (-1)=}
=
12
−
2
+
0
−
0
−
8
+
3
=
10
−
8
+
3
=
5.
{\displaystyle =12-2+0-0-8+3=10-8+3=5.}
Determinantą galima surasti ir kitu budu, pridėjus antrą determinanto stulpelį, padaugintą iš 3, prie pirmo stulpelio:
d
=
|
A
|
=
|
3
−
1
0
−
2
1
1
2
−
1
4
|
=
|
0
−
1
0
1
1
1
−
1
−
1
4
|
=
(
−
1
)
⋅
(
−
1
)
1
+
2
|
1
1
−
1
4
|
=
1
⋅
4
−
1
⋅
(
−
1
)
=
5.
{\displaystyle d=|A|={\begin{vmatrix}3&-1&0\\-2&1&1\\2&-1&4\end{vmatrix}}={\begin{vmatrix}0&-1&0\\1&1&1\\-1&-1&4\end{vmatrix}}=(-1)\cdot (-1)^{1+2}{\begin{vmatrix}1&1\\-1&4\end{vmatrix}}=1\cdot 4-1\cdot (-1)=5.}
Randame matricos A visus adjunktus:
A
11
=
(
−
1
)
1
+
1
|
1
1
−
1
4
|
=
5
;
A
12
=
(
−
1
)
1
+
2
|
−
2
1
2
4
|
=
10
;
A
13
=
(
−
1
)
1
+
3
|
−
2
1
2
−
1
|
=
0
;
{\displaystyle A_{11}=(-1)^{1+1}{\begin{vmatrix}1&1\\-1&4\end{vmatrix}}=5;\qquad A_{12}=(-1)^{1+2}{\begin{vmatrix}-2&1\\2&4\end{vmatrix}}=10;\qquad A_{13}=(-1)^{1+3}{\begin{vmatrix}-2&1\\2&-1\end{vmatrix}}=0;}
A
21
=
(
−
1
)
2
+
1
|
−
1
0
−
1
4
|
=
4
;
A
22
=
(
−
1
)
2
+
2
|
3
0
2
4
|
=
12
;
A
23
=
(
−
1
)
2
+
3
|
3
−
1
2
−
1
|
=
(
−
1
)
(
−
3
−
(
−
2
)
)
=
1
;
{\displaystyle A_{21}=(-1)^{2+1}{\begin{vmatrix}-1&0\\-1&4\end{vmatrix}}=4;\qquad A_{22}=(-1)^{2+2}{\begin{vmatrix}3&0\\2&4\end{vmatrix}}=12;\qquad A_{23}=(-1)^{2+3}{\begin{vmatrix}3&-1\\2&-1\end{vmatrix}}=(-1)(-3-(-2))=1;}
A
31
=
(
−
1
)
3
+
1
|
−
1
0
1
1
|
=
−
1
;
A
32
=
(
−
1
)
3
+
2
|
3
0
−
2
1
|
=
−
3
;
A
33
=
(
−
1
)
3
+
3
|
3
−
1
−
2
1
|
=
1.
{\displaystyle A_{31}=(-1)^{3+1}{\begin{vmatrix}-1&0\\1&1\end{vmatrix}}=-1;\qquad A_{32}=(-1)^{3+2}{\begin{vmatrix}3&0\\-2&1\end{vmatrix}}=-3;\qquad A_{33}=(-1)^{3+3}{\begin{vmatrix}3&-1\\-2&1\end{vmatrix}}=1.}
Toliau sudarome ir apskaičiuojame atvirkštinę A matricą:
A
−
1
=
1
|
A
|
⋅
[
A
11
A
21
A
31
A
12
A
22
A
32
A
13
A
23
A
33
]
=
1
5
⋅
[
5
4
−
1
10
12
−
3
0
1
1
]
=
[
1
4
5
−
1
5
2
12
5
−
3
5
0
1
5
1
5
]
.
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{bmatrix}}={\frac {1}{5}}\cdot {\begin{bmatrix}5&4&-1\\10&12&-3\\0&1&1\end{bmatrix}}={\begin{bmatrix}1&{\frac {4}{5}}&-{\frac {1}{5}}\\2&{\frac {12}{5}}&-{\frac {3}{5}}\\0&{\frac {1}{5}}&{\frac {1}{5}}\end{bmatrix}}.}
Sudauginę A matricą su jos atvirkštine matrica
A
−
1
{\displaystyle A^{-1}}
, gauname vienetinę matricą:
E
=
A
A
−
1
=
[
3
−
1
0
−
2
1
1
2
−
1
4
]
⋅
[
1
4
5
−
1
5
2
12
5
−
3
5
0
1
5
1
5
]
=
[
3
⋅
1
+
(
−
1
)
⋅
2
+
0
⋅
0
3
⋅
4
5
+
(
−
1
)
⋅
12
5
+
0
⋅
1
5
3
⋅
(
−
1
5
)
+
(
−
1
)
⋅
(
−
3
5
)
+
0
⋅
1
5
−
2
⋅
1
+
1
⋅
2
+
1
⋅
0
−
2
⋅
4
5
+
1
⋅
12
5
+
1
⋅
1
5
−
2
⋅
(
−
1
5
)
+
1
⋅
(
−
3
5
)
+
1
⋅
1
5
2
⋅
1
+
(
−
1
)
⋅
2
+
4
⋅
0
2
⋅
4
5
+
(
−
1
)
⋅
12
5
+
4
⋅
1
5
2
⋅
(
−
1
5
)
+
(
−
1
)
⋅
(
−
3
5
)
+
4
⋅
1
5
]
=
{\displaystyle E=AA^{-1}={\begin{bmatrix}3&-1&0\\-2&1&1\\2&-1&4\end{bmatrix}}\cdot {\begin{bmatrix}1&{\frac {4}{5}}&-{\frac {1}{5}}\\2&{\frac {12}{5}}&-{\frac {3}{5}}\\0&{\frac {1}{5}}&{\frac {1}{5}}\end{bmatrix}}={\begin{bmatrix}3\cdot 1+(-1)\cdot 2+0\cdot 0&3\cdot {\frac {4}{5}}+(-1)\cdot {\frac {12}{5}}+0\cdot {\frac {1}{5}}&3\cdot (-{\frac {1}{5}})+(-1)\cdot (-{\frac {3}{5}})+0\cdot {\frac {1}{5}}\\-2\cdot 1+1\cdot 2+1\cdot 0&-2\cdot {\frac {4}{5}}+1\cdot {\frac {12}{5}}+1\cdot {\frac {1}{5}}&-2\cdot (-{\frac {1}{5}})+1\cdot (-{\frac {3}{5}})+1\cdot {\frac {1}{5}}\\2\cdot 1+(-1)\cdot 2+4\cdot 0&2\cdot {\frac {4}{5}}+(-1)\cdot {\frac {12}{5}}+4\cdot {\frac {1}{5}}&2\cdot (-{\frac {1}{5}})+(-1)\cdot (-{\frac {3}{5}})+4\cdot {\frac {1}{5}}\end{bmatrix}}=}
=
[
3
−
2
+
0
12
5
−
12
5
+
0
−
3
5
+
3
5
+
0
−
2
+
2
+
0
−
8
5
+
12
5
+
1
5
2
5
−
3
5
+
1
5
2
−
2
+
0
8
5
−
12
5
+
4
5
−
2
5
+
3
5
+
4
5
]
=
[
1
0
0
0
1
0
0
0
1
]
,
{\displaystyle ={\begin{bmatrix}3-2+0&{\frac {12}{5}}-{\frac {12}{5}}+0&-{\frac {3}{5}}+{\frac {3}{5}}+0\\-2+2+0&-{\frac {8}{5}}+{\frac {12}{5}}+{\frac {1}{5}}&{\frac {2}{5}}-{\frac {3}{5}}+{\frac {1}{5}}\\2-2+0&{\frac {8}{5}}-{\frac {12}{5}}+{\frac {4}{5}}&-{\frac {2}{5}}+{\frac {3}{5}}+{\frac {4}{5}}\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}},}