Atvirkštinė matrica
A
−
1
{\displaystyle A^{-1}}
yra tokia matricos
A
=
[
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
]
{\displaystyle A={\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}}}
matrica, kad
A
A
−
1
=
A
−
1
A
=
E
;
{\displaystyle AA^{-1}=A^{-1}A=E;}
čia E yra vienetinė matrica (EA = AE = A ).
Atvirkštinė matrica gaunama taip:
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}}.}
|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=det\;A=|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}}{\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}},}
Antros eilės matricos atvirkštinė matrica[ keisti ]
Tegu duota antros eilės (kvadratinė) matrica
A
=
[
a
1
a
2
a
3
a
4
]
.
{\displaystyle A={\begin{bmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{bmatrix}}.}
Jos atvirkštinė matrica apskaičiuojama taip:
A
−
1
=
1
|
A
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
;
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}};}
čia |A| yra matricos A determinantas:
d
=
|
A
|
=
|
a
1
a
2
a
3
a
4
|
=
a
1
a
4
−
a
2
a
3
.
{\displaystyle d=|A|={\begin{vmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{vmatrix}}=a_{1}a_{4}-a_{2}a_{3}.}
Parodysime kaip ši antros eilės matricos atvirkštinės matricos formulė buvo gauta (ji gauta pagal tuos pačius dėsnius kaip ir trečios ir aukštesnės eilės atvirkštinės matricos).
Matricos
A
=
[
a
11
a
12
a
21
a
22
]
{\displaystyle A={\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}}
adjunktai yra tokie:
A
11
=
(
−
1
)
1
+
1
a
22
=
a
22
,
A
12
=
(
−
1
)
1
+
2
a
21
=
−
a
21
,
{\displaystyle A_{11}=(-1)^{1+1}a_{22}=a_{22},\;\;A_{12}=(-1)^{1+2}a_{21}=-a_{21},}
A
21
=
(
−
1
)
2
+
1
a
12
=
−
a
12
,
A
22
=
(
−
1
)
2
+
2
a
11
=
a
11
.
{\displaystyle A_{21}=(-1)^{2+1}a_{12}=-a_{12},\;\;A_{22}=(-1)^{2+2}a_{11}=a_{11}.}
Toliau kaip ir trečios eilės matricai, sukeičiame adjunktų eilutes su stulpeliais (kolonom) ir sudedam į kvadratinę matricą padalintą iš matricos A determinanto:
A
−
1
=
1
|
A
|
⋅
[
A
11
A
21
A
12
A
22
]
=
1
|
A
|
⋅
[
a
22
−
a
12
−
a
21
a
11
]
.
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}}={\frac {1}{|A|}}\cdot {\begin{bmatrix}a_{22}&-a_{12}\\-a_{21}&a_{11}\end{bmatrix}}.}
Gavome atvirkštinę matricos A matricą
A
−
1
.
{\displaystyle A^{-1}.}
A
=
[
3
5
1
2
]
.
{\displaystyle A={\begin{bmatrix}3&5\\1&2\end{bmatrix}}.}
Jos determinantas yra lygus:
d
=
|
A
|
=
|
3
5
1
2
|
=
3
⋅
2
−
5
⋅
1
=
1.
{\displaystyle d=|A|={\begin{vmatrix}3&5\\1&2\end{vmatrix}}=3\cdot 2-5\cdot 1=1.}
Matricos A atvirkštinė matrica yra tokia:
A
−
1
=
1
|
A
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
1
⋅
[
2
−
5
−
1
3
]
=
[
2
−
5
−
1
3
]
.
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{1}}\cdot {\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}={\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}.}
Sudaugtinę matricą A su jos atvirkštine matrica
A
−
1
{\displaystyle A^{-1}}
gausime vienetinę matricą E :
A
A
−
1
=
[
3
5
1
2
]
[
2
−
5
−
1
3
]
=
[
3
⋅
2
+
5
⋅
(
−
1
)
3
⋅
(
−
5
)
+
5
⋅
3
1
⋅
2
+
2
⋅
(
−
1
)
1
⋅
(
−
5
)
+
2
⋅
3
]
=
[
1
0
0
1
]
=
E
.
{\displaystyle AA^{-1}={\begin{bmatrix}3&5\\1&2\end{bmatrix}}{\begin{bmatrix}2&-5\\-1&3\end{bmatrix}}={\begin{bmatrix}3\cdot 2+5\cdot (-1)&3\cdot (-5)+5\cdot 3\\1\cdot 2+2\cdot (-1)&1\cdot (-5)+2\cdot 3\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}=E.}
A
=
[
a
1
a
2
a
3
a
4
]
=
[
8
4
3
2
]
.
{\displaystyle A={\begin{bmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{bmatrix}}={\begin{bmatrix}8&4\\3&2\end{bmatrix}}.}
Jos determinantas yra lygus:
d
=
|
A
|
=
|
8
4
3
2
|
=
8
⋅
2
−
4
⋅
3
=
16
−
12
=
4.
{\displaystyle d=|A|={\begin{vmatrix}8&4\\3&2\end{vmatrix}}=8\cdot 2-4\cdot 3=16-12=4.}
Matricos A atvirkštinė matrica yra tokia:
A
−
1
=
1
|
A
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
4
⋅
[
2
−
4
−
3
8
]
=
[
1
2
−
1
−
3
4
2
]
.
{\displaystyle A^{-1}={\frac {1}{|A|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{4}}\cdot {\begin{bmatrix}2&-4\\-3&8\end{bmatrix}}={\begin{bmatrix}{\frac {1}{2}}&-1\\-{\frac {3}{4}}&2\end{bmatrix}}.}
Sudaugtinę matricą A su jos atvirkštine matrica
A
−
1
{\displaystyle A^{-1}}
gausime vienetinę matricą E :
A
A
−
1
=
[
8
4
3
2
]
[
1
2
−
1
−
3
4
2
]
=
[
8
⋅
1
2
+
4
⋅
(
−
3
4
)
8
⋅
(
−
1
)
+
4
⋅
2
3
⋅
1
2
+
2
⋅
(
−
3
4
)
3
⋅
(
−
1
)
+
2
⋅
2
]
=
[
4
−
3
−
8
+
8
3
2
+
(
−
3
2
)
−
3
+
4
]
=
[
1
0
0
1
]
=
E
.
{\displaystyle AA^{-1}={\begin{bmatrix}8&4\\3&2\end{bmatrix}}{\begin{bmatrix}{\frac {1}{2}}&-1\\-{\frac {3}{4}}&2\end{bmatrix}}={\begin{bmatrix}8\cdot {\frac {1}{2}}+4\cdot (-{\frac {3}{4}})&8\cdot (-1)+4\cdot 2\\3\cdot {\frac {1}{2}}+2\cdot (-{\frac {3}{4}})&3\cdot (-1)+2\cdot 2\end{bmatrix}}={\begin{bmatrix}4-3&-8+8\\{\frac {3}{2}}+(-{\frac {3}{2}})&-3+4\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}=E.}
U
1
=
[
a
1
a
2
a
3
a
4
]
=
[
2
i
−
i
2
]
.
{\displaystyle U_{1}={\begin{bmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{bmatrix}}={\begin{bmatrix}{\sqrt {2}}&i\\-i&{\sqrt {2}}\end{bmatrix}}.}
Šią matricą transponavus ir jos kompleksinius skaičius pakeitus sujungtiniais, gaunama ta pati matrica. Tai yra, matrica
U
1
{\displaystyle U_{1}}
yra hermitinė, nes
U
1
∗
=
U
1
.
{\displaystyle U_{1}^{*}=U_{1}.}
Rasime jos atvirkštinę matricą
U
1
−
1
.
{\displaystyle U_{1}^{-1}.}
Jos determinantas yra lygus:
d
=
|
U
1
|
=
|
2
i
−
i
2
|
=
2
⋅
2
−
i
⋅
(
−
i
)
=
2
+
i
2
=
2
−
1
=
1.
{\displaystyle d=|U_{1}|={\begin{vmatrix}{\sqrt {2}}&i\\-i&{\sqrt {2}}\end{vmatrix}}={\sqrt {2}}\cdot {\sqrt {2}}-i\cdot (-i)=2+i^{2}=2-1=1.}
Matricos
U
1
{\displaystyle U_{1}}
atvirkštinė matrica yra tokia:
U
1
−
1
=
1
|
U
1
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
1
⋅
[
2
−
i
i
2
]
=
[
2
−
i
i
2
]
.
{\displaystyle U_{1}^{-1}={\frac {1}{|U_{1}|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{1}}\cdot {\begin{bmatrix}{\sqrt {2}}&-i\\i&{\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}{\sqrt {2}}&-i\\i&{\sqrt {2}}\end{bmatrix}}.}
Sudaugtinę matricą
U
1
{\displaystyle U_{1}}
su jos atvirkštine matrica
U
1
−
1
{\displaystyle U_{1}^{-1}}
gauname:
U
1
U
1
−
1
=
[
2
i
−
i
2
]
[
2
−
i
i
2
]
=
[
2
⋅
2
+
i
2
2
⋅
(
−
i
)
+
i
⋅
2
−
i
2
+
i
2
−
i
⋅
(
−
i
)
+
2
⋅
2
]
=
[
2
−
1
−
i
2
+
i
2
0
i
2
+
2
]
=
[
1
0
0
1
]
=
E
.
{\displaystyle U_{1}U_{1}^{-1}={\begin{bmatrix}{\sqrt {2}}&i\\-i&{\sqrt {2}}\end{bmatrix}}{\begin{bmatrix}{\sqrt {2}}&-i\\i&{\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}{\sqrt {2}}\cdot {\sqrt {2}}+i^{2}&{\sqrt {2}}\cdot (-i)+i\cdot {\sqrt {2}}\\-i{\sqrt {2}}+i{\sqrt {2}}&-i\cdot (-i)+{\sqrt {2}}\cdot {\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}2-1&-i{\sqrt {2}}+i{\sqrt {2}}\\0&i^{2}+2\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}=E.}
Matrica
U
1
{\displaystyle U_{1}}
nėra unitarinė matrica (unitarinėms matricoms galioja tokia lygybė:
U
∗
=
U
−
1
{\displaystyle U^{*}=U^{-1}}
), nes
U
1
∗
≠
U
1
−
1
.
{\displaystyle U_{1}^{*}\neq U_{1}^{-1}.}
Bet Aukštosios Algebros vadovėlyje teigiama, kad matrica
U
1
{\displaystyle U_{1}}
yra unitarinė (būna klaidų ir vadovėliuose).
U
2
=
[
a
1
a
2
a
3
a
4
]
=
[
2
i
5
5
−
2
i
]
.
{\displaystyle U_{2}={\begin{bmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{bmatrix}}={\begin{bmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{bmatrix}}.}
Matrica
U
2
{\displaystyle U_{2}}
nėra hermitinė nes
U
2
∗
≠
U
2
.
{\displaystyle U_{2}^{*}\neq U_{2}.}
Rasime matricos
U
2
{\displaystyle U_{2}}
atvirkštinę matricą
U
2
−
1
.
{\displaystyle U_{2}^{-1}.}
Matricos
U
2
{\displaystyle U_{2}}
determinantas yra lygus:
d
=
|
U
2
|
=
|
2
i
5
5
−
2
i
|
=
2
i
⋅
(
−
2
i
)
−
5
⋅
5
=
−
4
i
2
−
5
=
4
−
5
=
−
1.
{\displaystyle d=|U_{2}|={\begin{vmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{vmatrix}}=2i\cdot (-2i)-{\sqrt {5}}\cdot {\sqrt {5}}=-4i^{2}-5=4-5=-1.}
Matricos
U
2
{\displaystyle U_{2}}
atvirkštinė matrica yra tokia:
U
2
−
1
=
1
|
U
2
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
−
1
⋅
[
−
2
i
−
5
−
5
2
i
]
=
[
2
i
5
5
−
2
i
]
.
{\displaystyle U_{2}^{-1}={\frac {1}{|U_{2}|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{-1}}\cdot {\begin{bmatrix}-2i&-{\sqrt {5}}\\-{\sqrt {5}}&2i\end{bmatrix}}={\begin{bmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{bmatrix}}.}
Matome, kad
U
2
=
U
2
−
1
,
{\displaystyle U_{2}=U_{2}^{-1},}
bet
U
2
∗
≠
U
2
−
1
,
{\displaystyle U_{2}^{*}\neq U_{2}^{-1},}
nes
U
2
∗
=
[
−
2
i
5
5
2
i
]
.
{\displaystyle U_{2}^{*}={\begin{bmatrix}-2i&{\sqrt {5}}\\{\sqrt {5}}&2i\end{bmatrix}}.}
Vadovėlyje sakoma, kad kadangi determinantas
|
U
|
{\displaystyle |U|}
unitarinės matricos dažnai yra kompleksinis skaičius, tai tik jo modulis yra lygus 1 (kompleksinio skaičiaus
a
+
b
i
{\displaystyle a+bi}
modulis yra
a
2
+
b
2
{\displaystyle {\sqrt {a^{2}+b^{2}}}}
). Ir dar sakoma, kad unitarinėms ir ortogonalinėms matricoms
|
U
|
2
=
1.
{\displaystyle |U|^{2}=1.}
Bet ir padauginus matricą
[
−
2
i
−
5
−
5
2
i
]
{\displaystyle {\begin{bmatrix}-2i&-{\sqrt {5}}\\-{\sqrt {5}}&2i\end{bmatrix}}}
ne iš -1, o iš
(
−
1
)
2
=
1
,
{\displaystyle (-1)^{2}=1,}
vis tiek negaunama, kad
U
2
∗
{\displaystyle U_{2}^{*}}
yra lygu
U
2
−
1
.
{\displaystyle U_{2}^{-1}.}
Sudaugtinę matricą
U
2
{\displaystyle U_{2}}
su jos atvirkštine matrica
U
2
−
1
{\displaystyle U_{2}^{-1}}
gauname:
U
2
U
2
−
1
=
[
2
i
5
5
−
2
i
]
[
2
i
5
5
−
2
i
]
=
[
(
2
i
)
2
+
5
⋅
5
2
i
5
+
5
⋅
(
−
2
i
)
5
⋅
2
i
+
(
−
2
i
)
⋅
5
(
5
)
2
+
(
−
2
i
)
2
]
=
[
−
4
+
5
2
i
5
−
2
i
5
2
i
5
−
2
i
5
5
−
4
]
=
[
1
0
0
1
]
=
E
.
{\displaystyle U_{2}U_{2}^{-1}={\begin{bmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{bmatrix}}{\begin{bmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{bmatrix}}={\begin{bmatrix}(2i)^{2}+{\sqrt {5}}\cdot {\sqrt {5}}&2i{\sqrt {5}}+{\sqrt {5}}\cdot (-2i)\\{\sqrt {5}}\cdot 2i+(-2i)\cdot {\sqrt {5}}&({\sqrt {5}})^{2}+(-2i)^{2}\end{bmatrix}}={\begin{bmatrix}-4+5&2i{\sqrt {5}}-2i{\sqrt {5}}\\2i{\sqrt {5}}-2i{\sqrt {5}}&5-4\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}=E.}
Kur ten mato unitarines matricas Aukštosios Algebros vadovėlio pavyzdyje? Nes matricos
U
1
{\displaystyle U_{1}}
ir
U
2
{\displaystyle U_{2}}
nėra unitarinės.
Surasime matricų
U
1
{\displaystyle U_{1}}
ir
U
2
{\displaystyle U_{2}}
sandaugos atvirkštinę matricą.
U
3
=
U
1
U
2
=
[
2
i
−
i
2
]
[
2
i
5
5
−
2
i
]
=
[
(
2
2
+
5
)
i
2
+
10
2
+
10
−
(
2
2
+
5
)
i
]
.
{\displaystyle U_{3}=U_{1}U_{2}={\begin{bmatrix}{\sqrt {2}}&i\\-i&{\sqrt {2}}\end{bmatrix}}{\begin{bmatrix}2i&{\sqrt {5}}\\{\sqrt {5}}&-2i\end{bmatrix}}={\begin{bmatrix}(2{\sqrt {2}}+{\sqrt {5}})i&2+{\sqrt {10}}\\2+{\sqrt {10}}&-(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}.}
Surandame matricos
U
3
{\displaystyle U_{3}}
determinantą:
|
U
3
|
=
|
(
2
2
+
5
)
i
2
+
10
2
+
10
−
(
2
2
+
5
)
i
|
=
−
(
2
2
+
5
)
2
i
2
−
(
2
+
10
)
2
=
(
2
2
+
5
)
2
−
(
2
+
10
)
2
=
(
8
+
4
10
+
5
)
−
(
4
+
4
10
+
10
)
=
13
+
4
10
−
14
−
4
10
=
−
1.
{\displaystyle |U_{3}|={\begin{vmatrix}(2{\sqrt {2}}+{\sqrt {5}})i&2+{\sqrt {10}}\\2+{\sqrt {10}}&-(2{\sqrt {2}}+{\sqrt {5}})i\end{vmatrix}}=-(2{\sqrt {2}}+{\sqrt {5}})^{2}i^{2}-(2+{\sqrt {10}})^{2}=(2{\sqrt {2}}+{\sqrt {5}})^{2}-(2+{\sqrt {10}})^{2}=(8+4{\sqrt {10}}+5)-(4+4{\sqrt {10}}+10)=13+4{\sqrt {10}}-14-4{\sqrt {10}}=-1.}
Matricos
U
3
{\displaystyle U_{3}}
atvirkštinė matrica yra tokia:
U
3
−
1
=
1
|
U
3
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
−
1
⋅
[
−
(
2
2
+
5
)
i
−
(
2
+
10
)
−
(
2
+
10
)
(
2
2
+
5
)
i
]
=
[
(
2
2
+
5
)
i
(
2
+
10
)
(
2
+
10
)
−
(
2
2
+
5
)
i
]
.
{\displaystyle U_{3}^{-1}={\frac {1}{|U_{3}|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{-1}}\cdot {\begin{bmatrix}-(2{\sqrt {2}}+{\sqrt {5}})i&-(2+{\sqrt {10}})\\-(2+{\sqrt {10}})&(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}={\begin{bmatrix}(2{\sqrt {2}}+{\sqrt {5}})i&(2+{\sqrt {10}})\\(2+{\sqrt {10}})&-(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}.}
Matome, kad
U
3
=
U
3
−
1
.
{\displaystyle U_{3}=U_{3}^{-1}.}
Jei matrica
U
3
{\displaystyle U_{3}}
yra unitarinė, tai pritaikius jai žvaigždutinę operaciją (*), turėtume gauti atvirkštinę matricą
U
3
−
1
.
{\displaystyle U_{3}^{-1}.}
Žiūrim:
U
3
∗
=
[
−
(
2
2
+
5
)
i
(
2
+
10
)
(
2
+
10
)
(
2
2
+
5
)
i
]
{\displaystyle U_{3}^{*}={\begin{bmatrix}-(2{\sqrt {2}}+{\sqrt {5}})i&(2+{\sqrt {10}})\\(2+{\sqrt {10}})&(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}}
(žvaigždutinė operacija sukeičia elementų eilutes su stulpeliais (transponuoja) ir kompleksinius skaičius pakeičia jiems jungtiniais kompleksiniais skaičiais). Gavome, kad matrica
U
3
{\displaystyle U_{3}}
nėra unitarinė (nes
U
3
∗
≠
U
3
−
1
{\displaystyle U_{3}^{*}\neq U_{3}^{-1}}
) ir nėra ermitinė (nes
U
3
∗
≠
U
3
{\displaystyle U_{3}^{*}\neq U_{3}}
). Vadovėlyje teigiama, kad matrica
U
3
{\displaystyle U_{3}}
yra unitarinė (ir kad dviejų unitarinių matricų sandauga yra unitarinė matrica), bet taip nėra.
Sudauginę matricą
U
3
{\displaystyle U_{3}}
su jos atvirkštine matrica
U
3
−
1
{\displaystyle U_{3}^{-1}}
gauname:
U
3
U
3
−
1
=
[
(
2
2
+
5
)
i
2
+
10
2
+
10
−
(
2
2
+
5
)
i
]
[
(
2
2
+
5
)
i
2
+
10
2
+
10
−
(
2
2
+
5
)
i
]
=
[
(
2
2
+
5
)
2
i
2
+
(
2
+
10
)
2
(
2
2
+
5
)
(
2
+
10
)
i
−
(
2
2
+
5
)
(
2
+
10
)
i
(
2
2
+
5
)
(
2
+
10
)
i
−
(
2
2
+
5
)
(
2
+
10
)
i
(
2
+
10
)
2
+
[
−
(
2
2
+
5
)
i
]
2
]
=
{\displaystyle U_{3}U_{3}^{-1}={\begin{bmatrix}(2{\sqrt {2}}+{\sqrt {5}})i&2+{\sqrt {10}}\\2+{\sqrt {10}}&-(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}{\begin{bmatrix}(2{\sqrt {2}}+{\sqrt {5}})i&2+{\sqrt {10}}\\2+{\sqrt {10}}&-(2{\sqrt {2}}+{\sqrt {5}})i\end{bmatrix}}={\begin{bmatrix}(2{\sqrt {2}}+{\sqrt {5}})^{2}i^{2}+(2+{\sqrt {10}})^{2}&(2{\sqrt {2}}+{\sqrt {5}})(2+{\sqrt {10}})i-(2{\sqrt {2}}+{\sqrt {5}})(2+{\sqrt {10}})i\\(2{\sqrt {2}}+{\sqrt {5}})(2+{\sqrt {10}})i-(2{\sqrt {2}}+{\sqrt {5}})(2+{\sqrt {10}})i&(2+{\sqrt {10}})^{2}+[-(2{\sqrt {2}}+{\sqrt {5}})i]^{2}\end{bmatrix}}=}
=
[
−
(
8
+
4
10
+
5
)
+
(
4
+
4
10
+
10
)
0
0
(
4
+
4
10
+
10
)
−
(
8
+
4
10
+
5
)
]
=
[
−
(
13
+
4
10
)
+
(
14
+
4
10
)
0
0
(
14
+
4
10
)
−
(
13
+
4
10
)
]
=
[
1
0
0
1
]
=
E
.
{\displaystyle ={\begin{bmatrix}-(8+4{\sqrt {10}}+5)+(4+4{\sqrt {10}}+10)&0\\0&(4+4{\sqrt {10}}+10)-(8+4{\sqrt {10}}+5)\end{bmatrix}}={\begin{bmatrix}-(13+4{\sqrt {10}})+(14+4{\sqrt {10}})&0\\0&(14+4{\sqrt {10}})-(13+4{\sqrt {10}})\end{bmatrix}}={\begin{bmatrix}1&0\\0&1\end{bmatrix}}=E.}
Unitariųjų matricų pavyzdžiai[ keisti ]
Vokiškoj, berdos, Vikipedijoje yra teisingų unitariųjų matricų pavyzdžių: https://de.wikipedia.org/wiki/Unitäre_Matrix
U
=
(
0
i
i
0
)
.
{\displaystyle U={\begin{pmatrix}0&i\\i&0\end{pmatrix}}.}
Jos atvirkštinė matrica yra tokia (matricos U determinantas lygus 1):
U
−
1
=
(
0
−
i
−
i
0
)
=
U
∗
.
{\displaystyle U^{-1}={\begin{pmatrix}0&-i\\-i&0\end{pmatrix}}=U^{*}.}
Sudauginę matricą
U
−
1
{\displaystyle U^{-1}}
su U gauname:
U
−
1
U
=
(
0
−
i
−
i
0
)
⋅
(
0
i
i
0
)
=
(
−
i
2
0
0
−
i
2
)
=
(
1
0
0
1
)
=
E
.
{\displaystyle U^{-1}\,U={\begin{pmatrix}0&-i\\-i&0\end{pmatrix}}\cdot {\begin{pmatrix}0&i\\i&0\end{pmatrix}}={\begin{pmatrix}-i^{2}&0\\0&-i^{2}\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}=E.}
U
=
1
2
(
1
+
i
1
−
i
1
−
i
1
+
i
)
.
{\displaystyle U={\frac {1}{2}}{\begin{pmatrix}1+i&1-i\\1-i&1+i\end{pmatrix}}.}
Jos determinantas yra (skaičiuojant determinantą iš kiekvienos matricos eilutės (arba stulpelio) reikia iškelti dauginamąjį, todėl 1/4, o ne 1/2 prieš determinantą):
|
U
|
=
1
4
|
1
+
i
1
−
i
1
−
i
1
+
i
|
=
1
4
[
(
1
+
i
)
2
−
(
1
−
i
)
2
]
=
1
4
[
1
+
2
i
+
i
2
−
(
1
−
2
i
+
i
2
)
]
=
1
4
(
2
i
+
2
i
)
=
i
.
{\displaystyle |U|={\frac {1}{4}}{\begin{vmatrix}1+i&1-i\\1-i&1+i\end{vmatrix}}={\frac {1}{4}}[(1+i)^{2}-(1-i)^{2}]={\frac {1}{4}}[1+2i+i^{2}-(1-2i+i^{2})]={\frac {1}{4}}(2i+2i)=i.}
Pritaikę žvaigždutinę operaciją gauname tokią matricą:
U
∗
=
1
2
(
1
−
i
1
+
i
1
+
i
1
−
i
)
.
{\displaystyle U^{*}={\frac {1}{2}}{\begin{pmatrix}1-i&1+i\\1+i&1-i\end{pmatrix}}.}
O sudauginę ką tik gautą matricą su matrica U gauname:
U
∗
U
=
1
2
(
1
−
i
1
+
i
1
+
i
1
−
i
)
⋅
1
2
(
1
+
i
1
−
i
1
−
i
1
+
i
)
=
1
4
(
2
(
1
−
i
)
(
1
+
i
)
(
1
−
i
)
2
+
(
1
+
i
)
2
(
1
+
i
)
2
+
(
1
−
i
)
2
2
(
1
+
i
)
(
1
−
i
)
)
=
(
1
0
0
1
)
=
E
.
{\displaystyle U^{*}\,U={\frac {1}{2}}{\begin{pmatrix}1-i&1+i\\1+i&1-i\end{pmatrix}}\cdot {\frac {1}{2}}{\begin{pmatrix}1+i&1-i\\1-i&1+i\end{pmatrix}}={\frac {1}{4}}{\begin{pmatrix}2(1-i)(1+i)&(1-i)^{2}+(1+i)^{2}\\(1+i)^{2}+(1-i)^{2}&2(1+i)(1-i)\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}=E.}
Matricos U atvirkštinė matrica yra tokia:
U
−
1
=
1
|
U
|
⋅
[
a
4
−
a
2
−
a
3
a
1
]
=
1
2
i
⋅
[
1
+
i
−
(
1
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=
{\displaystyle U^{-1}={\frac {1}{|U|}}\cdot {\begin{bmatrix}a_{4}&-a_{2}\\-a_{3}&a_{1}\end{bmatrix}}={\frac {1}{2i}}\cdot {\begin{bmatrix}1+i&-(1-i)\\-(1-i)&1+i\end{bmatrix}}={\frac {1}{2i}}\cdot {\begin{bmatrix}1+i&-1+i\\-1+i&1+i\end{bmatrix}}=}
=
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+
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[
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+
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+
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.
{\displaystyle ={\frac {i}{2i\cdot i}}\cdot {\begin{bmatrix}1+i&-1+i\\-1+i&1+i\end{bmatrix}}={\frac {-i}{2}}\cdot {\begin{bmatrix}1+i&-1+i\\-1+i&1+i\end{bmatrix}}={\frac {1}{2}}\cdot {\begin{bmatrix}-i(1+i)&-i(-1+i)\\-i(-1+i)&-i(1+i)\end{bmatrix}}={\frac {1}{2}}\cdot {\begin{bmatrix}-i+1&i+1\\i+1&-i+1\end{bmatrix}}={\frac {1}{2}}\cdot {\begin{bmatrix}1-i&1+i\\1+i&1-i\end{bmatrix}}.}
Visiškai teisingai. Gavome, kad
U
∗
=
U
−
1
.
{\displaystyle U^{*}=U^{-1}.}
Matrica U be jokių abejonių yra unitarioji (unitarinė ~1960 metų vadovėlyje).