|
tartibli determinant
deyiladi.
Demak,
ta’rifga ko‘ra,
77
66
15
53
41
22
34
a
a
a
a
a
a
a
ko‘paytma
7-tartibli
determinantning hadi bo‘lishi mumkin.
7
6
3
1
4
2
5
7
6
5
4
3
2
1
7
6
5
3
1
2
4
7
6
1
5
4
2
3
F
F
7
0
0
2
3
1
1
2
invP
;
1
)
1
(
)
1
(
)
(
7
2
invP
F
h
, bundan
77
66
15
53
41
22
34
a
a
a
a
a
a
a
ko‘paytma 7-tartibli
determinantning hadi va uning ishorasi manfiy ekani kelib chiqadi. ◄
Tuzuvchi: dots. S.S.Sadaddinova
1 – tоpshiriq.
Ikkitа А vа B mаtritsаlаr bеrilgаn.
а
) 3А─2B+5E ifodani hisoblang;
b) АB va BА ko‘paytmalarni hisoblang va kommutativlikka tekshiring;
c)
T
T
T
A
B
AB
tenglikni tekshiring:
1.0. А =
4
1
4
6
0
3
1
3
5
, B=
1
4
5
3
3
1
1
2
2
.
1.1. А =
2
4
3
6
7
8
3
1
2
, B =
1
2
1
4
5
3
2
1
2
.
1.2. А =
1
1
3
3
4
2
6
5
3
, B =
3
5
4
0
1
3
5
8
2
.
1.3. А =
1
0
1
1
1
2
1
1
2
, B =
1
2
1
6
4
2
0
6
3
.
1.4. А =
2
4
0
5
2
9
11
1
6
, B =
2
3
1
7
2
0
1
0
3
.
1.5. А =
1
2
1
2
0
1
2
1
3
, B =
1
7
3
1
1
2
2
1
0
.
1.6. А =
3
1
4
1
3
1
2
3
2
, B=
0
3
5
2
1
3
1
2
3
.
1.7. А =
1
2
2
0
1
3
3
7
6
, B =
7
3
4
2
1
4
5
0
2
.
1.8. А =
2
2
1
4
1
3
4
3
2
, B =
2
9
1
2
6
0
1
3
3
.
Tuzuvchi: dots. S.S.Sadaddinova
1.9. А =
1
1
0
2
3
1
1
6
2
, B =
3
2
3
5
0
4
2
3
4
.
1.10. А =
4
1
4
6
0
3
5
3
1
, B=
1
2
5
3
3
1
1
2
1
.
1.11. А =
7
1
10
1
1
1
4
9
6
, B =
2
5
0
3
4
3
1
1
1
.
1.12. А =
8
1
2
7
1
3
3
0
1
, B=
4
6
5
1
0
3
4
5
3
.
1.13. А =
1
4
8
1
3
1
2
1
5
, B =
1.14. А =
4
3
4
6
3
3
5
2
2
, B =
1
2
1
3
3
2
1
1
1
.
1.15. А =
4
3
4
6
0
3
5
2
1
, B=
1
2
1
3
3
2
1
1
1
.
1.16. А =
5
0
3
4
2
1
2
4
5
, B =
2
2
1
1
7
3
5
1
5
.
1.17. А =
7
2
2
2
3
4
0
1
3
, B =
1
6
1
1
3
5
0
7
2
.
1.18. А =
2
3
0
1
1
5
5
1
1
8
, B =
2
0
1
1
2
3
5
2
3
.
1.19. А =
3
2
4
3
8
1
2
7
3
, B =
5
1
2
1
4
2
3
5
0
.
0
6
1
2
1
7
5
5
3
Tuzuvchi: dots. S.S.Sadaddinova
1.20. А =
5
7
4
1
5
3
0
1
3
, B =
2
0
3
4
8
1
2
0
1
.
1.21. А =
7
1
10
1
1
1
4
9
6
, B =
2
5
0
3
4
3
1
1
1
.
1.22. А =
8
1
2
7
1
3
3
0
1
, B=
4
6
5
1
0
3
4
5
3
.
1.23. А =
1
4
8
1
3
1
2
1
5
, B =
1.24. А =
4
3
4
6
3
3
5
2
2
, B =
1
2
1
3
3
2
1
1
1
.
1.25. А =
4
3
4
6
0
3
5
2
1
, B=
1
2
1
3
3
2
1
1
1
.
1.26. А =
5
0
3
4
2
1
2
4
5
, B =
2
2
1
1
7
3
5
1
5
.
1.27. А =
7
2
2
2
3
4
0
1
3
, B =
1
6
1
1
3
5
0
7
2
.
1.28. А =
2
3
0
1
1
5
5
1
1
8
, B =
2
0
1
1
2
3
5
2
3
.
1.29. А =
3
2
4
3
8
1
2
7
3
, B =
5
1
2
1
4
2
3
5
0
.
1.30. А =
5
7
4
1
5
3
0
1
3
, B =
2
0
3
4
8
1
2
0
1
.
0
6
1
2
1
7
5
5
3
Tuzuvchi: dots. S.S.Sadaddinova
1.31. А =
3
2
4
3
8
0
2
7
3
, B =
5
1
2
1
4
2
3
5
0
.
1.32. А =
5
7
4
1
5
3
0
1
9
, B =
2
0
3
4
8
1
2
0
7
.
1.33. А =
7
1
7
1
1
1
4
9
0
, B =
2
5
5
3
4
3
1
0
1
.
1.34. А =
8
1
2
7
1
3
3
0
1
, B=
4
6
5
1
0
3
4
5
11
.
2 - tоpshiriq.
Bеrilgаn Δ dеtеrminаnt uchun
a
i2
, a
3j
elеmеntlаrning minоrlаri vа аlgеbrаik
to‘ldiruvchilаrini tоping. Δ dеtеrminаntni
а)
i-
sаtr elеmеntlаri bo‘yichа yoyib;
b)
j-
ustun elеmеntlаri bo‘yichа yoyib;
c)
j-
ustun elеmеntlаrini nollаrgа аylаntirib hisоblаng.
2.0.
1
2
1
3
4
1
2
2
6
0
3
2
1
4
2
1
i=
4 , j = 3.
2.1.
14
-
5
3
2
4
6
0
1
5
2
-
6
3
0
2
-
1
1
. 2.2.
6
0
2
4
3
1
-
2
0
0
9
-
3
6
3
1
-
0
2
.
i =
4 , j = 1 .
i =
1 , j = 3 .
2.3.
3
1
5
0
2
0
4
3
0
1
1
1
1
2
7
2
2.4.
8
6
4
2
3
1
3
5
2
8
2
3
5
1
5
4
i=
4 , j = 1 .
i=
1 , j = 3 .
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