Matritsalar va ular ustida amallar. Matritsalarni ko’paytirish, teskari matritsani topish. Matritsaning rangi

Matritsalar 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.

Matritsalar 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

Matritsalar 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.

Matritsalar 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.
Misol 1. 𝐴 matritsa berilgan:
2 −4 1
a) 𝐴 =^{−1 2}; b) 𝐴 =1 −5 3.
1 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.

a) 𝐴 =^{−1 2}matritsaning determinantini hisoblaymiz: