Reja: Matritsalar va ular ustida amallar

M atritsalar va ular ustida amallar

𝐴𝐵 ko’paytmaning mavjudligidan 𝐵𝐴 ko’paytmaning mavjudligi kelib chiqmaydi. 𝐴𝐵 va 𝐵𝐴 ko’paytmalar mavjud bo’lgan taqdirda ham, odatda (ko’p hollarda), 𝐴𝐵 va 𝐵𝐴 ko’paytmalar bir-biriga teng bo’lmaydi: 𝐴𝐵 ≠ 𝐵𝐴. Agar 𝐴𝐵 = 𝐵𝐴 bo’lsa, u holda 𝐴 va 𝐵 matritsalar o’zaro o’rin almashinuvchi (kommutativ) matritsalar deyiladi.
Ma’lumki, har doim 𝐴𝐵 𝐶 = 𝐴 𝐵𝐶 tenglik o’rinli.

M atritsalar va ular ustida amallar

Misol 1. 𝐴𝐵 va 𝐵𝐴 ko’paytmalarni toping.
4 −5 8
𝐴 =,
1 3 −1

−1 5
𝐵 =−2 −3.

3. 
   4
   

𝐴𝐵 ko’paytmani topamiz:
−1 5

3. 
   −5 8
   

𝐴𝐵 =−2 −3=
1 3 −1

3. 
   4
   
4. 
   ⋅ (−1) + (−5) ⋅ (−2) + 8 ⋅ 3 4 ⋅ 5 + (−5) ⋅ (−3) + 8 ⋅ 430 67
   

==.
1 ⋅ (−1) + 3 ⋅ (−2) + (−1) ⋅ 3 1 ⋅ 5 + 3 ⋅ (−3) + (−1) ⋅ 4−10 −8

M atritsalar va ular ustida amallar

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

M atritsalar va ular ustida amallar

Misol 2. 𝐴𝐵 va 𝐵𝐴 ko’paytmalarni toping.
3 51 −5
𝐴 =, 𝐵 =.
1 2−1 2
Hisoblaymiz:
3 51 −53 ⋅ 1 + 5 ⋅ (−1) 3 ⋅ (−5) + 5 ⋅ 2−2 −5
𝐴𝐵 ===,
1 2−1 21 ⋅ 1 + 2 ⋅ (−1) 1 ⋅ (−5) + 2 ⋅ 2−1 −1
1 −53 51 ⋅ 3 + (−5) ⋅ 1 1 ⋅ 5 + (−5) ⋅ 2−2 −5
𝐵𝐴 ===.
−1 21 2(−1) ⋅ 3 + 2 ⋅ 1 (−1) ⋅ 5 + 2 ⋅ 2−1 −1
Shunday qilib, 𝐴𝐵 = 𝐵𝐴 ekan.

1. 
   3
   

2
𝐴 =−1 1, 𝐵 =
1

2. 
   5
   

− 1
−6 1 , 𝐶 =2.

3. 
   −1
   

4

Ko’paytmalarni hisoblaymiz:
5 3 −2
𝐴𝐵 =−1 9 −2

9 3 −3
− 7
𝐴𝐵 𝐶 = 11 ,
−15

− 10
𝐵𝐶 = , 𝐴
1
ya`ni𝐴𝐵𝐶 = 𝐴𝐵𝐶.
− 7
𝐵𝐶 = 11 ,
−15

𝑛 − tartibli kvadrat matritsa berilgan bo’lsin:
𝑎11 𝑎12 . . . 𝑎1𝑛
𝐴 =𝑎.21. . 𝑎.22. . .. .. .. 𝑎.2.𝑛.
𝑎𝑛1 𝑎𝑛2 . . . 𝑎𝑛𝑛

Agar 𝐴 matritsaning determinanti noldan farqli
𝑎11 𝑎12 . . . 𝑎1𝑛
𝑑𝑒𝑡 𝐴 =𝑎.21. . 𝑎.22. . .. .. .. 𝑎.2.𝑛.≠ 0
𝑎𝑛1 𝑎𝑛2 . . . 𝑎𝑛𝑛
bo’lsa, 𝐴 matritsa aynimagan matritsa deyiladi. Agar 𝑑𝑒𝑡 𝐴 = 0 bo’lsa, 𝐴 matritsa aynigan matritsa deyiladi.
𝐴 matritsaga teskari matritsa 𝐴⁻¹ ko’rinishda belgilanadi. Teskari matritsa tushunchasi faqat aynimagan kvadrat matritsalarga taalluqlidir. Ushbu
1 0 . . . 0
0 1 . . . 0
𝐸 =
. . . . . . . . . . . .

0 0 . . . 1
kvadrat matritsa birlik matritsa deyiladi.
Ushbu
𝑎11 𝑎21 . . . 𝑎𝑛1
𝐴𝑇 =𝑎.12. . 𝑎.22. . .. .. .. 𝑎.𝑛2. .
𝑎1𝑛 𝑎2𝑛 . . . 𝑎𝑛𝑛
kvadrat matritsa 𝐴 matritsaga nisbatan transponirlangan matritsa deyiladi.
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:
𝐴11 𝐴21 . . . 𝐴𝑛1
_𝐴−1 ₌𝐴12 𝐴22 . . . 𝐴𝑛2_, . . . . . . . . . . . .
𝐴1𝑛 𝐴2𝑛 . . . 𝐴𝑛𝑛
bu yerda 𝐴_𝑖𝑗 − berilgan 𝐴 matritsaga nisbatan transponirlangan 𝐴^𝑇 matritsaning algebraik to’ldiruvchilari.