2-shaxsiy topshiriq.
Mаtritsаlаr vа dеtеrminаntlаr nаzаriyasi.
Chiziqli tеnglаmаlаr sistеmаsini yеchish
(A4, A5, A6
-
mashg‘ulotlarga doir bo‘lib, 5 ballik tizimda baholanadi
)
.
0-variantning bajarilish tartibi
1.0-misol.
1
2
1
3
3
1
0 7
2
3
-1
4
A
matritsa berilgan.
a) Ta’rifga ko‘ra matritsa rangini toping;
b) O‘rab turuvchi minorlar usulida
matritsa rangini aniqlang;
v) Elementar almashtirishlar yordamida matritsa rangini toping.
Yechilishi: ►
a) Matritsa rangini ta’rifga ko‘ra aniqlaymiz.
Matritsaning noldan farqli 1-tartibli minorlari mavjud.
Bizning matritsa
4
3
o‘lchamli, shuning uchun
bu matritsadan ajratish
mumkin bolgan 2-tartibli minorlar soni
18
6
3
2
4
2
3
С
С
ta.
Matritsaning noldan
farqli 2-tartibli minorlarini izlaymiz, bittasini topsak yetarli:
0
7
6
1
1
3
2
-
1
2
,
1
2
,
1
M
Matritsaning noldan farqli 3-tartibli minorlari soni
4
3
4
3
3
С
С
ta.
Ularni
tekshiramiz:
0
6
2
9
1
1
3
2
0
1
3
1
2
1
3
,
2
,
1
3
,
2
,
1
M
;
0
21
24
6
27
28
4
4
3
2
7
1
3
3
2
1
3
,
2
,
1
4
,
2
,
1
M
;
0
7
12
9
14
4
1
2
7
0
3
3
1
1
3
,
2
,
1
4
,
3
,
1
M
;
0
14
4
3
21
4
1
3
7
0
1
3
1
2
4
,
3
,
2
4
,
3
,
2
M
Ko‘rish mumkinki, 3-tartibli minorlarning hammasi nolga teng. Demak,
matritsaning noldan farqli eng yuqori tartibli minorining tartibi 2 ga teng. Shuning
uchun matritsa
rangi
2
)
(
A
rank
.
b) Endi o‘rab turuvchi minorlar usulida matritsa rangini aniqlaymiz:
1
2
1
3
3
1
0 7
2
3
-1
4
A
Aytaylik, matritsaning
1
11
a
elementini olamiz. Uni o‘rab turgan, ya’ni o‘z
ichiga olgan
0
7
6
1
1
3
2
-
1
2
,
1
2
,
1
M
matritsani noldan farqli ekanini aniqladik.
Boshqa 2-tartibli minorlarni tekshirib o‘tirishning hojati yo’q. Endi shu matritsani
o‘z ichiga olgan 3-tartibli minorlarni tekshiramiz:
0
6
2
9
1
1
3
2
0
1
3
1
2
1
3
,
2
,
1
3
,
2
,
1
M
;
0
21
24
6
27
28
4
4
3
2
7
1
3
3
2
1
3
,
2
,
1
4
,
2
,
1
M
;
.
0
7
12
9
14
4
1
2
7
0
3
3
1
1
3
,
2
,
1
4
,
3
,
1
M
Bundan ko‘rinadiki, matritsaning barcha o‘rab turuvchi
3-tartibli minorlari nolga
teng. U holda
2
)
(
A
rank
.
v) Matritsa rangini elementar almashtirishlar yordamida aniqlash uchun
berilgan matritsa ustida elementar almashtirishlar bajarib, pog‘onasimon
matritsaga keltiramiz:
1
2
1
3
1
2
1
3
1
2
1
3
3
1
0 7
0
7
-3 2
0
7
-3 2 .
2
3
-1 4
0
7
-3
2
0
~
0
~
0
0
Matritsa pog‘onasimon matritsaga keltirildi. 3-satrning barcha elementlari
nollardan iborat boʻlganligi sababli, uni tashlab yuboramiz. U holda noldan farqli
eng katta tartibli minorning tartibi 2 ga teng:
0
7
7
0
2
1
1
M
Shuning uchun berilgan
A
matritsa uchun
2
)
(
A
rank
. ◄
2.0-misol.
1
2
1
3
4
1
2
2
6
0
3
2
1
4
2
1
A
ga А
1
teskari matritsani
a)
algebraik to‘ldiruvchilar
yordamida;
b)
elementar almashtirishlar yordamida
toping va
1
1
A
AA
E
A
tenglikni tekshiring:
Yechilishi: ► a) Algebraik to‘ldiruvchilar yordamida teskari matritsani
topamiz,
buning uchun quyidagi algoritm bo’yicha ish yuritamiz:
1)
Matritsaning determinantini hisoblaymiz:
390
1
2
1
3
4
1
2
2
6
0
3
2
1
4
2
1
2) Har bir elementning algebraik to‘ldiruvchisini topamiz, ular 16 ta:
57
24
0
6
24
0
3
1
2
1
4
1
2
6
0
3
)
1
(
11
1
1
11
M
A
;
42
)
12
16
6
4
24
12
(
4
2
2
6
3
2
1
2
1
)
1
(
43
3
4
43
M
A
;
39
0
4
24
16
0
3
1
2
2
0
3
2
4
2
1
)
1
(
44
4
4
44
M
A
;
3) Algebraik to‘ldiruvchilardan matritsa tuzamiz va uni transponirlaymiz. Hosil
bo‘lgan matritsa berilgan matritsaga qo‘shma matritsa bo‘ladi:
39
42
24
81
26
38
96
66
39
18
66
21
13
86
12
57
B
39
26
39
13
42
38
18
86
24
96
66
12
81
66
21
57
T
B
4)
T
B
A
1
1
formuladan foydalanib,
A
matritsaga teskari matritsani topamiz:
39
26
39
13
42
38
18
86
24
96
66
12
81
66
21
57
390
1
1
1
T
B
A
.
5)
Topilgan
1
A
matirtsa to‘g‘ri topilganini bilish uchun
E
A
A
1
formula
yordamida tekshirib ko‘ramiz:
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
2
1
3
4
1
2
2
6
0
3
2
1
4
2
1
39
26
39
13
42
38
18
86
24
96
66
12
81
66
21
57
390
1
1
E
A
A
.
