−
−
=
−
2
1
2
1
2
1
0
1
2
0
0
1
1
A
.
Teskari matritsaga ega bo‘lamiz.
Teskari matritsaning asosiy xossalari.
;
)
1
1
1
−
−
=
A
A
( )
;
)
2
1
1
A
A
=
−
−
(
)
;
)
3
1
1
1
−
−
−
=
A
B
B
A
( )
( )
=
−
−
1
1
)
4
A
A
.
3-xossaning isbotini ko‘ramiz:
( )
(
) (
)
( )
1
1
1
1
1
1
AB B A
A BB
A
AE A
AA
E
−
−
−
−
−
−
=
=
=
=
,
(
)
( )
(
)
( )
1
1
1
1
1
1
B A
AB
B A A B
BE B
BB
E
−
−
−
−
−
−
=
=
=
=
bundan
( )
1
1
1
−
−
−
=
AB
A
B
.
4-xossaning isbotini ko‘ramiz:
( ) ( )
E
E
A
A
A
A
=
=
=
−
−
1
1
,
bundan
( )
( )
1
1
−
−
=
A
A
.
5-ta’rif.
Agar
A
kvadrat
matritsa uchun
t
t
A A
A A
E
=
=
( ya’ni
1
t
A
A
−
=
)
bo‘lsa, u holda
A
matritsa orthogonal matritsa deyiladi.
3-tеорема.
Har qanday orthogonal matritsa uchun teskari matritsa mavjud va u
ham orthogonal matritsa bo‘ladi.
Bu teorema
( )
A
A
=
bolganidan ,
E
A
A
A
A
=
=
kelib chiqadi.
4-tеорема.
Orthogonal matritsalarning
ko‘paytmasi
ham
orthogonal matritsa
bo‘ladi.
3. Ekvivalent almashtirishlar yordamida teskari matritsani hisoblash.
Teskari matritsani topishning Gauss-Jordan usulida maxsusmas matritsani shu
tartibdagi birlik
matritsa bilan kengaytiriladi, kengaytirilgan matritsa satrlari ustida
elementar almashtirish to kengaytirilgan matritsa birinchi
qismida birlik matritsa
hosil boʻlguncha olib boriladi, natijada kengaytirilgan matritsaning ikkinchi qismida
"Science and Education" Scientific Journal
August 2021 / Volume 2 Issue 8
www.openscience.uz
298
berilgan matritsaga teskari boʻlgan matritsa hosil boʻladi. Bu jarayonni Gauss-Jordan
modifikatsiyasi (yoki formulasi) koʻrinishida yozishimiz mumkin:
( )
(
)
1
~
A E
E A
−
4-misol.
Gauss-Jordan usulida berilgan matritsaga teskari matritsani toping.
1 1
1
1 2
1 .
2 2
4
A
=
−
Yechish.
(
)
3 6
o‘lchamli
(
)
/
Г
A E
=
kengaytirilgan matritsani yozamiz. Avval
matritsaning satrlari ustida elementar almashtirishlar bajarib uni pog‘onasimon
ko‘rinishga
keltiramiz
(
)
1
1
/
Г
A B
=
, keyin
(
)
1
2
/
Г
E A
−
=
ko‘rinishga keltiramiz.
1
1
1 1
0
0
1
2
1 0
1
0
2
2
4 0
0
1
2
Г
II I
III
I
=
−
−
−
1
1
1
1 1
0
0
0
1
2 1 1
0
0
0
2
2
0
1
Г
II III
=
− −
+
−
1
1
1 1
0
0
0
1
0 3
1
1
0
0
2 2
0
1
2
III
−
−
1
1
1 1
0
0
0
1
0 3
1
1
0
0
1
1
1 0
2
I
II
III
− −
−
−
2
3
5
1
1
0
0
2
0
1
0 3
1
1
0
0
1
1
1
0
2
Г
−
−
−
=
−
Demak,
1
3
5
1
2
3
1
1
.
1
1
0
2
A
−
−
−
= −
−
Tekshiramiz:
1
3
5
1
1
1
1
1
0
0
2
1
2
1
3
1
1
0
1
0 .
2
2
4
1
0
0
1
1
0
2
AA
−
−
−
=
−
−
=
−
"Science and Education" Scientific Journal
August 2021 / Volume 2 Issue 8
www.openscience.uz
299