Mustaqil ish amaliy Matematika fanidan

Matritsalar va ular ustida amallar

퐵퐴 ko’paytmani topamiz:
퐵퐴 =
−1 5
−2 −3
3 4
⋅
4 −5 8
1 3 −1
=
(−1) ⋅ 4 + 5 ⋅ 1
(−2) ⋅ 4 + (−3) ⋅ 1
3 ⋅ 4 + 4 ⋅ 1
(−1) ⋅ (−5) + 5 ⋅ 3
(−2) ⋅ (−5) + (−3) ⋅ 3
3 ⋅ (−5) + 4 ⋅ 3
(−1) ⋅ 8 + 5 ⋅ (−1)
(−2) ⋅ 8 + (−3) ⋅ (−1)
3 ⋅ 8 + 4 ⋅ (−1)
=
1 20 −13
−11 1 −13
16 −3 20
.
Shunday qilib, 퐴퐵 ≠ 퐵퐴 ekan.

Matritsalar va ular ustida amallar

Misol 2. 퐴퐵 va 퐵퐴 ko’paytmalarni toping.
퐴 =
3 5
1 2
, 퐵 =
1 −5
−1 2
.
Hisoblaymiz:
퐴퐵 =
⋅
3 5 1 −5
1 2 −1 2
=
3 ⋅ 1 + 5 ⋅ (−1)
1 ⋅ 1 + 2 ⋅ (−1)
3 ⋅ (−5) + 5 ⋅ 2
1 ⋅ (−5) + 2 ⋅ 2
=
−2 −5
−1 −1
,
퐵퐴 =
⋅
1 −5 3 5
−1 2 1 2
=
1 ⋅ 3 + (−5) ⋅ 1
(−1) ⋅ 3 + 2 ⋅ 1
1 ⋅ 5 + (−5) ⋅ 2
(−1) ⋅ 5 + 2 ⋅ 2
=
−2 −5
−1 −1
.
Shunday qilib, 퐴퐵 = 퐵퐴 ekan.

Matritsalar va ular ustida amallar

Misol 3. 퐴퐵 퐶 va 퐴 퐵퐶 ko’paytmalarni toping.
퐴 =
1 3
−1 1
2 5
, 퐵 =
2 −6 1
1 3 −1
, 퐶 =
−1
2
4
.
Ko’paytmalarni hisoblaymiz:

	5	3	−2			−7
퐴퐵 =	−1	9	−2	,	퐴퐵 퐶 =	11	,
	9	3	−3			−15

퐵퐶 =
−10
1
, 퐴 퐵퐶 =
−7
11
−15
,
ya`ni
퐴퐵 퐶 = 퐴 퐵퐶 .

Teskari matritsa

푛 − tartibli kvadrat matritsa berilgan bo’lsin:
퐴 =
푎11
푎21
. . .
푎푛1
푎12
푎22
. . .
푎푛2
. . . 푎1푛
. . . 푎2푛
. . . . . .
. . . 푎푛푛
Agar 퐴 matritsaning determinanti noldan farqli
푑푒푡 퐴 =
푎11
푎21
. . .
푎푛1
푎12
푎22
. . .
푎푛2
. . . 푎1푛
. . . 푎2푛
. . . . . .
. . . 푎푛푛
≠ 0
bo’lsa, 퐴 matritsa aynimagan matritsa deyiladi. Agar 푑푒푡 퐴 = 0 bo’lsa, 퐴 matritsa
aynigan matritsa deyiladi.

Teskari matritsa

퐴 matritsaga teskari matritsa 퐴−1 ko’rinishda belgilanadi. Teskari matritsa tushunchasi faqat aynimagan kvadrat matritsalarga taalluqlidir.
Ushbu
퐸 =

1	0 . . . 0
0	1 . . . 0
. . .	. . . . . . . . .
0	0 . . . 1

퐴푇 =
kvadrat matritsa birlik matritsa deyiladi. Ushbu
푎11
푎12
. . .
푎1푛
푎21
푎22
. . .
푎2푛
. . . 푎푛1
. . . 푎푛2
. . . . . .
. . . 푎푛푛
kvadrat matritsa 퐴 matritsaga nisbatan transponirlangan matritsa deyiladi.

Teskari matritsa

Aynimagan 퐴 matritsa berilgan bo’lsin. Agar
퐴 ⋅ 퐴−1 = 퐴−1 ⋅ 퐴 = 퐸
bo’lsa, 퐴−1 matritsa 퐴 matritsaga teskari matritsa deyiladi.
퐴 matritsaga teskari 퐴−1 matritsani topish formulasi:
퐴−1 =
1
푑푒푡 퐴
퐴11
퐴12
. . .
퐴1푛
퐴21
퐴22
. . .
퐴2푛
. . . 퐴푛1
. . . 퐴푛2
. . . . . .
. . . 퐴푛푛
,
bu yerda 퐴푖푗 − berilgan 퐴 matritsaga nisbatan transponirlangan 퐴푇 matritsaning algebraik to’ldiruvchilari.

Teskari matritsa

Misol 1. 퐴 matritsa berilgan:
a) 퐴 =
−1 2
1 3
; b) 퐴 =
2 −4 1
1 −5 3
1 −1 1
.
퐴 matritsa aynimagan matritsa ekanligiga ishonch hosil qiling, 퐴 matritsaga
teskari 퐴−1 matritsani toping va 퐴 ⋅ 퐴−1 = 퐴−1 ⋅ 퐴 = 퐸 tengliklarning bajarilishini tekshiring.

Teskari matritsa

a) 퐴 =
−1 2
1 3
matritsaning determinantini hisoblaymiz:
푑푒푡 퐴 =
−1 2
1 3
= (−1) ⋅ 3 − 2 ⋅ 1 = −5 ≠ 0.
퐴 matritsaning aynimagan matritsa ekanligiga ishonch hosil qildik. Endi algebraik to’ldiruvchilarni topamiz:
퐴11 = 3, 퐴12 = −1, 퐴21 = −2, 퐴22 = −1.
Teskari matritsani yoza olamiz:
퐴−1 = − 1
3 −2
5 −1 −1
=
−3/5 2/5
1/5 1/5
.

Teskari matritsa

퐴 ⋅ 퐴−1 = 퐴−1 ⋅ 퐴 = 퐸
tengliklarning bajarilishini tekshirib ko’ramiz.
퐴 ⋅ 퐴−1 =
−1 2
1 3
⋅
−3/5 2/5
1/5 1/5
=
3
5
+ 2 ⋅
1
5
2
5
(−1) ⋅ + 2 ⋅
1
5
(−1) ⋅ −
1 ⋅ −
3
5
+ 3 ⋅
1
5
2
5
1 ⋅ + 3 ⋅
1
5
=
1 0
0 1
= 퐸;
퐴−1 ⋅ 퐴 =
−3/5 2/5
1/5 1/5
⋅
−1 2
1 3
=
3
2
2 3
− ⋅ (−1) + ⋅ 1 − ⋅ 2 + ⋅ 3
5 5 5 5
⋅ (−1) + ⋅ 1

1		1		1		1
5		5		5		5

⋅ 2 + ⋅ 3
=
1 0
0 1
= 퐸.

Teskari matritsa

b) 퐴 =
2 −4 1
1 −5 3
1 −1 1
matritsaning determinantini hisoblaymiz:
푑푒푡 퐴 =
2 −4 1
1 −5 3
1 −1 1
= −8 ≠ 0.
퐴 matritsaning aynimagan matritsa ekanligiga ishonch hosil qildik. Endi algebraik to’ldiruvchilarni topamiz:
퐴11 = −2, 퐴12 = 2, 퐴13 = 4, 퐴21 = 3,
퐴22 = 1, 퐴23 = −2, 퐴31 = −7, 퐴32 = −5, 퐴33 = −6.
Teskari matritsani yoza olamiz: 퐴−1 = − 1
8
−2 3 −7
2 1 −5
4 −2 −6
.

Teskari matritsa

퐴 ⋅ 퐴−1 = 퐴−1 ⋅ 퐴 = 퐸 tengliklarning bajarilishini tekshirib ko’ramiz.
퐴−1 ⋅ 퐴 = −
1
8

−2	3	−7		2	−4	1
2	1	−5	⋅	1	−5	3	=
4	−2	−6		1	−1	1

= −
1
8
−2 ⋅ −4 + 3 ⋅ −5 + −7 ⋅ −1
2 ⋅ −4 + 1 ⋅ −5 + −5 ⋅ −1
−2 ⋅ 1 + 3 ⋅ 3 +
2 ⋅ 1 + 1 ⋅ 3 +
−2 ⋅ 2 + 3 ⋅ 1 + −7 ⋅ 1
2 ⋅ 2 + 1 ⋅ 1 + −5 ⋅ 1
4 ⋅ 2 + −2 ⋅ 1 + −6 ⋅ 1
4 ⋅ −4 + −2 ⋅ −5 +
−6 ⋅ −1 4 ⋅ 1 + −2 ⋅ 3 + −6 ⋅ 1
−7 ⋅ 1
−5 ⋅ 1 =

−8	0	0		1	0	0
0	−8	0	=	0	1	0	= 퐸;
0	0	−8		0	0	1

1
= −
8
Xuddi shu kabi 퐴 ⋅ 퐴−1 = 퐸 ekanligini ko’rsatish mumkin.

Matritsaning rangi

Matritsaning rangi tushunchasini kiritamiz. 퐴 matritsada 푘 ta satrlar va 푘 ta usunlarni ajratamiz, bu yerda 푘 soni 푚 va 푛 sonlarining kichigidan ham kichik yoki teng
(푘 ≤ 푚푖푛 푚, 푛 ). Ajratib olingan 푘 ta satrlar va 푘 ta usunlarning kesishmasida turgan elementlardan tuzilgan 푘 −tartibli determinant matritsadan yaralgan minor yoki
determinant deyiladi. Masalan,
7 −1 4 5
1 8 1 3
4 −2 0 −6
matritsa berilgan bo’lsin.

Matritsaning rangi

푘 = 2 bo’lganda
7 −1 1 3
1 8 , 0 −6
,
,
−1 5 8 1 4 5
−2 −6 −2 0 , 1 3
determinantlar berilgan matritsadan yaralgan determinantlardir.
퐴 matritsadan yaralgan determinantlar ichidan noldan farqlilarini ajratib olamiz. Ana shu noldan farqli determinantlar tartibining eng kattasi 푨 matritsaning rangi deyiladi (푟푎푛푔퐴 deb belgilanadi).
Agar 퐴 matritsadan yaralgan 푘 −tartibli determinantlarning hammasi nolga teng bo’lsa, u holda 푟푎푛푔퐴 < 푘 bo’ladi.

Matritsaning rangi

Teorema 1. Quyidagi elementar (oddiy) almashtirishlar bajarilganda matritsaning rangi o’zgarmaydi:

• Ixtiyoriy ikkita parallel qatorlarning o’rinlari almashtirilganda;
• Qatorning har bir elementini bir xil 휆 ≠ 0 songa ko’paytirilganda;
• Qatorning elementlariga ixtiyoriy boshqa qatorning mos elementlarini bir xil songa ko’paytirib qo’shganda.

Matritsaning rangi

Agar biror matritsa boshqa matritsadan elementar almashtirishlar yordamida hosil qilinsa, bunday matritsalar ekvivalent matritsalar deyiladi.
퐴 va 퐵 matritsalarning ekvivalentligi 퐴 ∼ 퐵 deb belgilanadi.
Tartibi berilgan matritsaning rangiga teng bo’lgan noldan farqli har qanday minor
matritsaning bazis minori deyiladi.

Matritsaning rangini topish usullari

1	1	2	3	−1
2	−1	0	−4	−5
−1	−1	0	−3	−2
6	3	4	8	−3

Birlar va nollar usuli.
Elementar almashtirishlar yordamida har qanday matritsani shunday ko’rinishga keltirish mumkinki, bunda matritsaning har bir qatori faqat nollardan yoki faqat nollardan va bitta birdan iborat bo’ladi. Hosil bo’lgan matritsa dastlabki matritsaga
ekvivalent bo’lganligi uchun oxirida qolgan birlarning soni dastlabki matritsaning rangi bo’ladi. Quyidagi matritsaning rangini va bazis minorini toping.
퐴 = .

Matritsaning rangini topish usullari

Yechish. 퐴 matritsaning uchinchi ustunini 1 ga ko’paytiramiz. So’ngra, hosil bo’lgan
2
birinchi satrni 2 ga ko’paytiramiz va uni to’rtinchi satrdan ayiramiz. Endi uchinchi
ustun uchta nollar va bitta birdan (birinchi satrda) iborat bo’ladi:
퐴 ∼
2 −1 0 −4 −5
∼

1

1

1

3

−1

2

2

2

6

−2

2 −1 0 −4 −5
∼
∼

−1	−1	0	−3	−2	−1	−1	0	−3	−2
6	3	2	8	−3	6	3	2	8	−3
2	2	2	6	−2	2	2	1	6	−2

2	−1	0	−4	−5	∼	2	−1	0	−4	−5
−1	−1	0	−3	−2	−1		−1	0	−3	−2
4	1	0	2	−1	4		1	0	2	−1

.