Reja: Matritsalar va ular ustida amallar

Matritsalar va ular ustida amallar

𝐵𝐴 ko’paytmani topamiz:
𝐵𝐴 =
−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, 𝐴𝐵 ≠ 𝐵𝐴 ekan.

Matritsalar va ular ustida amallar

Misol 2. 𝐴𝐵 va 𝐵𝐴 ko’paytmalarni toping.
𝐴 =
3 5
1 2
, 𝐵 =
1 −5
−1 2
.
Hisoblaymiz:
𝐴𝐵 =
⋅
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
,
𝐵𝐴 =
⋅
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, 𝐴𝐵 = 𝐵𝐴 ekan.

Matritsalar va ular ustida amallar

Misol 3. 𝐴𝐵 𝐶 va 𝐴 𝐵𝐶 ko’paytmalarni toping.
𝐴 =
1 3
−1 1
2 5
, 𝐵 =
2 −6 1
1 3 −1
, 𝐶 =
−1
2
4
.
Ko’paytmalarni hisoblaymiz:

	5	3	−2			−7
𝐴𝐵 =	−1	9	−2	,	𝐴𝐵 𝐶 =	11	,
	9	3	−3			−15

𝐵𝐶 =
−10
1
, 𝐴 𝐵𝐶 =
−7
11
−15
,
ya`ni
𝐴𝐵 𝐶 = 𝐴 𝐵𝐶 .

Teskari matritsa

𝑛 − tartibli kvadrat matritsa berilgan bo’lsin:
𝐴 =
𝑎11
𝑎21
. . .
𝑎𝑛1
𝑎12
𝑎22
. . .
𝑎𝑛2
. . . 𝑎1𝑛
. . . 𝑎2𝑛
. . . . . .
. . . 𝑎𝑛𝑛
Agar 𝐴 matritsaning determinanti noldan farqli
𝑑𝑒𝑡 𝐴 =
𝑎11
𝑎21
. . .
𝑎𝑛1
𝑎12
𝑎22
. . .
𝑎𝑛2
. . . 𝑎1𝑛
. . . 𝑎2𝑛
. . . . . .
. . . 𝑎𝑛𝑛
≠ 0
bo’lsa, 𝐴 matritsa aynimagan matritsa deyiladi. Agar 𝑑𝑒𝑡 𝐴 = 0 bo’lsa, 𝐴 matritsa
aynigan matritsa deyiladi.

Teskari matritsa

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

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

𝐴𝑇 =
kvadrat matritsa birlik matritsa deyiladi. Ushbu
𝑎11
𝑎12
. . .
𝑎1𝑛
𝑎21
𝑎22
. . .
𝑎2𝑛
. . . 𝑎𝑛1
. . . 𝑎𝑛2
. . . . . .
. . . 𝑎𝑛𝑛
kvadrat matritsa 𝐴 matritsaga nisbatan transponirlangan matritsa deyiladi.

Teskari matritsa

Aynimagan 𝐴 matritsa berilgan bo’lsin. Agar
𝐴 ⋅ 𝐴−1 = 𝐴−1 ⋅ 𝐴 = 𝐸
bo’lsa, 𝐴−1 matritsa 𝐴 matritsaga teskari matritsa deyiladi.
𝐴 matritsaga teskari 𝐴−1 matritsani topish formulasi:
𝐴−1 =
1
𝑑𝑒𝑡 𝐴
𝐴11
𝐴12
. . .
𝐴1𝑛
𝐴21
𝐴22
. . .
𝐴2𝑛
. . . 𝐴𝑛1
. . . 𝐴𝑛2
. . . . . .
. . . 𝐴𝑛𝑛
,
bu yerda 𝐴𝑖𝑗 − berilgan 𝐴 matritsaga nisbatan transponirlangan 𝐴𝑇 matritsaning algebraik to’ldiruvchilari.

Teskari matritsa

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

Teskari matritsa

a) 𝐴 =
−1 2
1 3
matritsaning determinantini hisoblaymiz:
𝑑𝑒𝑡 𝐴 =
−1 2
1 3
= (−1) ⋅ 3 − 2 ⋅ 1 = −5 ≠ 0.
𝐴 matritsaning aynimagan matritsa ekanligiga ishonch hosil qildik. Endi algebraik to’ldiruvchilarni topamiz:
𝐴11 = 3, 𝐴12 = −1, 𝐴21 = −2, 𝐴22 = −1.
Teskari matritsani yoza olamiz:
𝐴−1 = − 1
3 −2
5 −1 −1
=
−3/5 2/5
1/5 1/5
.

Teskari matritsa

𝐴 ⋅ 𝐴−1 = 𝐴−1 ⋅ 𝐴 = 𝐸
tengliklarning bajarilishini tekshirib ko’ramiz.
𝐴 ⋅ 𝐴−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
= 𝐸;
𝐴−1 ⋅ 𝐴 =
−3/5 2/5
1/5 1/5
⋅
−1 2
1 3
=
3
5
3
5
2
2
− ⋅ (−1) + ⋅ 1 − ⋅ 2 + ⋅ 3
5 5
⋅ (−1) + ⋅ 1

1		1		1		1
5		5		5		5

⋅ 2 + ⋅ 3
=
1 0
0 1
= 𝐸.

Teskari matritsa

b) 𝐴 =
2 −4 1
1 −5 3
1 −1 1
matritsaning determinantini hisoblaymiz:
𝑑𝑒𝑡 𝐴 =
2 −4 1
1 −5 3
1 −1 1
= −8 ≠ 0.
𝐴 matritsaning aynimagan matritsa ekanligiga ishonch hosil qildik. Endi algebraik to’ldiruvchilarni topamiz:
𝐴11 = −2, 𝐴12 = 2, 𝐴13 = 4, 𝐴21 = 3,
𝐴22 = 1, 𝐴23 = −2, 𝐴31 = −7, 𝐴32 = −5, 𝐴33 = −6.
Teskari matritsani yoza olamiz: 𝐴−1 = − 1
8
−2 3 −7
2 1 −5
4 −2 −6
.

Teskari matritsa

𝐴 ⋅ 𝐴−1 = 𝐴−1 ⋅ 𝐴 = 𝐸 tengliklarning bajarilishini tekshirib ko’ramiz.
𝐴−1 ⋅ 𝐴 = −
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	= 𝐸;
0	0	−8		0	0	1

1
= −
8
Xuddi shu kabi 𝐴 ⋅ 𝐴−1 = 𝐸 ekanligini ko’rsatish mumkin.

Matritsaning rangi

Matritsaning rangi tushunchasini kiritamiz. 𝐴 matritsada 𝑘 ta satrlar va 𝑘 ta usunlarni ajratamiz, bu yerda 𝑘 soni 𝑚 va 𝑛 sonlarining kichigidan ham kichik yoki teng
(𝑘 ≤ 𝑚𝑖𝑛 𝑚, 𝑛 ). Ajratib olingan 𝑘 ta satrlar va 𝑘 ta usunlarning kesishmasida turgan elementlardan tuzilgan 𝑘 −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

𝑘 = 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.
𝐴 matritsadan yaralgan determinantlar ichidan noldan farqlilarini ajratib olamiz. Ana shu noldan farqli determinantlar tartibining eng kattasi 𝑨 matritsaning rangi deyiladi (𝑟𝑎𝑛𝑔𝐴 deb belgilanadi).
Agar 𝐴 matritsadan yaralgan 𝑘 −tartibli determinantlarning hammasi nolga teng bo’lsa, u holda 𝑟𝑎𝑛𝑔𝐴 < 𝑘 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 𝜆 ≠ 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.
𝐴 va 𝐵 matritsalarning ekvivalentligi 𝐴 ∼ 𝐵 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.
𝐴 = .

Matritsaning rangini topish usullari

Yechish. 𝐴 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:
𝐴 ∼
2 −1 0 −4 −5
∼

1

1

1

3

−1

2

2

2

6

−2

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

−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
−1 −1 0 −3 −2
4 1 0 2 −1
.