Reja: Matritsalar va ular ustida amallar

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:
1 3
−1 2
𝑑𝑒𝑡 𝐴 == (−1) ⋅ 3 − 2 ⋅ 1 = −5 ≠ 0.
1 3

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:
− 1 = − 1
𝐴
5

3. 
   − 2−3/5 2/5
   

=. −1 −11/5 1/5

𝐴 ⋅ 𝐴⁻¹ = 𝐴⁻¹ ⋅ 𝐴 = 𝐸
tengliklarning bajarilishini tekshirib ko’ramiz.
2
(−1) ⋅ + 2 ⋅
51 0
== 𝐸;
20 1

1 ⋅ + 3 ⋅
5
⋅ 2 +
−3/5 2/51 0
== 𝐸.
1 10 1
⋅ 2 +
5 5
1 −1 1
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.
−2 3 −7
Teskari matritsani yoza olamiz: 𝐴⁻¹ = − 2 1 −5 .

3. 
   −2 −6
   

𝐴 ⋅ 𝐴⁻¹ = 𝐴⁻¹ ⋅ 𝐴 = 𝐸 tengliklarning bajarilishini tekshirib ko’ramiz.
−2 3 −72 −4 1
𝐴⁻¹ ⋅ 𝐴 = −2 1 −51 −5 3= 4 −2 −61 −1 1

−2 ⋅ 2 + 3 ⋅ 1 + −7 ⋅ 1 −2 ⋅ −4 + 3 ⋅ −5 + −7 ⋅ −1 −2 ⋅ 1 + 3 ⋅ 3 + −7 ⋅ 1
= −2 ⋅ 2 + 1 ⋅ 1 + −5 ⋅ 1 2 ⋅ −4 + 1 ⋅ −5 + −5 ⋅ −1 2 ⋅ 1 + 1 ⋅ 3 + −5 ⋅ 1 =
4 ⋅ 2 + −2 ⋅ 1 + −6 ⋅ 1 4 ⋅ −4 + −2 ⋅ −5 + −6 ⋅ −1 4 ⋅ 1 + −2 ⋅ 3 + −6 ⋅ 1
−8 0 01 0 0
= −0 −8 00 1 0= 𝐸;
0 0 −80 0 1
Xuddi shu kabi 𝐴 ⋅ 𝐴⁻¹ = 𝐸 ekanligini ko’rsatish mumkin.

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.
𝑘 = 2 bo’lganda
7 −11 3−1 58 14 5
,,
1 80 −6−2 −6−2 01 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.