Chiziqli tenglamalar sistemasini yechish usullari. Reja

Yechish: Determinantni matritsa birinchi satr elementlari bo’ycha yoyib hisoblaymiz.

1 2 1
−2 3 0 3
0 −2

1 1
^|0 −2 3^{|=1· |} 3 1 1
|-2· |
3 1
|+1· |_{3 1} |=(-2· 1 − 1 · 3)-2(0· 1 − 3 ·

3)+(0· 1 + 3 · 2)=-5+18+6=19

Determinantning xossalari.

1. 
   xossa. Determinantning biror satr (ustuni) nollardan iborat bo’lsa, bu determinantning qiymati nolga teng.
   

^𝑎11 ^𝑎12 ^𝑎13

_[ 0 0 0
]=0

^𝑎31 ^𝑎32 ^𝑎33

2. 
   xossa. Determinantning qiymatini biror songa ko’paytirish uning biror satri
   

(ustuni) elementlarini shu songa ko’paytirishga teng kuchli.
^{𝑎11 𝑎12 𝑎13 𝑎}11 ^𝑎12 ^𝑎13
^[𝑘𝑎21 ^𝑘𝑎22 ^𝑘𝑎23^{] = 𝑘 [𝑎21 𝑎22 𝑎23]
𝑎}31 ^𝑎32 ^𝑎33 ^{𝑎31 𝑎32 𝑎33}

3. 
   xossa. Matritsani transponirlash natijasida uning determinantining qiymati
   

o’zgarmaydi , yani berilgan A matritsa uchun ^|𝐴^| = ^|𝐴^𝑇|.
^𝑎11 ^𝑎12 ^𝑎13 ^𝑎11 ^𝑎21 ^𝑎31
^[𝑎21 ^𝑎22 ^𝑎23^]=[𝑎12 ^𝑎22 ^𝑎32^]
𝑎31 ^𝑎32 ^𝑎33 ^𝑎13 ^𝑎23 ^𝑎33

4. 
   xossa. Determinantda ikkita satr yoki ikkita ustunning o’rni
   

almashtrilsa,determinantning qiymati o’z ishorasini almashtiradi.
^𝑎11 ^𝑎12 ^𝑎13 ^𝑎21 ^𝑎22 ^𝑎23
^[𝑎21 ^𝑎22 ^𝑎23^{]=− [𝑎}11 ^𝑎12 ^𝑎13^]
𝑎31 ^𝑎32 ^𝑎33 ^𝑎31 ^𝑎32 ^𝑎33

5. 
   xossa.Ikkita bir xil satrga (ustunga) ega bo’lgan determinant nolga teng.
   

^𝑎11 ^𝑎12 ^𝑎13
^[𝑎11 ^𝑎12 ^𝑎13^{] = 0
𝑎}31 ^𝑎32 ^𝑎33

6. 
   xossa.Determinantning ikkita satr (ustuni) o’zaro proportsional bo’lsa,bu determinantning qiymati nolga teng.
   

^𝑎11 ^𝑎12 ^𝑎13
^[𝑘𝑎11 ^𝑘𝑎12 ^𝑘𝑎13^{] = 0
𝑎}31 ^𝑎32 ^𝑎33

7. 
   xossa. Agar determinantning biror satri (ustuni) to’laligicha ikki qo’shiluvchining yig’indisidan iborat bo’lsa , uning qiymatini ikkita determinantlar
   

qiymatlari yig’indisi tarzida hisoblash mumkin.

^𝑎11 ^𝑎12 ^𝑎13
^𝑎11 ^𝑎21 ^𝑎31
^𝑎11 ^𝑎21 ^𝑎31

^[𝑎21 ^{+ 𝑏}1 ^𝑎22 ^{+ 𝑏}2 ^𝑎23 ^{+ 𝑏}3^{] = [𝑎12 𝑎22 𝑎32] + [ 𝑏}1 ^𝑏2 ^𝑏3 ^]

^𝑎31 ^𝑎32 ^𝑎33
^𝑎13 ^𝑎23 ^𝑎33
^𝑎13 ^𝑎23 ^𝑎33

8. 
   xossa. Agar determinantnng biror satri elementlarini bir xil songa ko’paytirib
   

boshqa biror satrga qo’shilsa uning qiymati o’zgarmaydi.
^{𝑎11 𝑎12 𝑎13 𝑎}11 ^𝑎12 ^𝑎13
^[𝑘𝑎11 ^{+ 𝑎}21 ^𝑘𝑎12 ^{+ 𝑎}22 ^𝑘𝑎13 ^{+ 𝑎}₂₃^{] = [𝑎21 𝑎22 𝑎23]
𝑎}31 ^𝑎32 ^𝑎33 ^{𝑎31 𝑎32 𝑎33}

9. 
   xossa. Matritsa biror satr elementlarini boshqa biror satr elementlarining algebraik to’ldiruvchilariga ko’paytmalarining yig’indisi nolga teng.
   

^𝑎11^𝐴31 ^{+ 𝑎}12^𝐴32^+𝑎13^𝐴33^{=0; 𝑎}21^𝐴31^+𝑎22^𝐴32^+𝑎23^𝐴33⁼⁰

10. 
    xossa. Kvadrat matritsalar ko’paytmasining determinant har bir matritsa determinantlari ko’paytmasiga teng,ya’ni ^|𝐴 · 𝐵^| = ^|𝐴^| · ^|𝐵^|
    

Ixtiyoriy A va B kvadrat matritsalar uchun det(AB)=detA·detB
Sonli jadvallar(matritsa,determinant) elementlari ustidadagi elementar almashtirishlar deganda satr elementlarini biror songa ko’paytirib boshqa bir satrga qo’shish ,satrlar o’rnini almashtirish tushuniladi.

Determinantlarni hisoblash.

1. 
   Misol. Berilgan matritsaning determinantini hisoblang.
   

2 2 −1
^A=( 2 −4 1 ⁾
−2 −2 2

1. 
   usul. Determinantni ixtiyoriy satri yoki ustuni elementlari bo’yicha yoyib hisoblash teoremasidan foydalanamiz. Masalan, determinantni 3-ustun elementlari bo’yicha
   

yoyamiz
:[ 2 2 −1]=−1⁽−1⁾¹⁺³ [ ^{2 −4}] + 1⁽−1⁾²⁺³ [ ^{2 2} ] +

2 −4 1
−2 −2 −2
−2 −2
−2 −2

2(−1)³⁺³ [^{2 2}
2 −4
] = −⁽−4 − 8⁾ − ⁽−4 + 4⁾ + 2⁽−8 − 4⁾ = 12 − 0 − 24 = −12

2. 
   usul. Determinantni hisoblashning uchburchak qoidasidan foydalanamiz:
   

2 2 −1
^[ 2 −4 1 ^{]=2· (−4) · 2 + 2 · 1 · (−2) + 2 · (−2) · (−1)— 1 · (−4) ·}
−2 −2 2
⁽−2⁾ −
2· 2 · 2 − ⁽−2⁾ · 1 · 2 = −16 − 4 + 4 − 8 + 8 + 4 = −12

3. 
   usul. Elementar almashtirishlar orqali determinantni biror satr yoki ustunni maksimal miqdorda nolli bo’lgan ko’rinishga keltiramiz va o’sha satr yoki ustun bo’yicha yoyamiz:
   

2 2 −1
^[ 2 −4 1
−2 −2 2
]=^{1 − 𝑠𝑎𝑡𝑟𝑔𝑎 3 − 𝑛i 𝑞𝑜′𝑠ℎ𝑎𝑚i𝑧^}=[
0 0 1
2 −4 1^]=
−2 −2 2

^{1 − 𝑠𝑎𝑡𝑟 𝑏𝑜^′𝑦i𝑐ℎ𝑎 𝑦𝑜𝑦𝑜𝑚i𝑧^}=1(−1)¹⁺³ [ ^{2 −4}]=-4-8=-12
−2 −2

4. 
   usul. Elementar almashtirishlar orqali uchburchak ko’rinishga keltiramz:
   

2 2 1

^[ 2 −4 1^{] {1 − 𝑠𝑎𝑡𝑟𝑛i − 1𝑔𝑎 𝑘𝑜′𝑝𝑎𝑦𝑡i𝑟i𝑏, 2 −}
−2 −2 2

2 2 −1

^{𝑠𝑎𝑡𝑟𝑔𝑎 𝑞𝑜′𝑠ℎ𝑎𝑚i𝑧𝑣𝑎 𝑛𝑎𝑡ij𝑎𝑛i 2 − 𝑠𝑎𝑡𝑟𝑔𝑎 𝑦𝑜𝑧𝑎𝑚i𝑧}=[} 0 −6 2 ^]=
−2 −2 2
^{1 − 𝑠𝑎𝑡𝑟𝑔𝑎 3 − 𝑛i 𝑞𝑜^′𝑠ℎ𝑎𝑚i𝑧 𝑣𝑎 𝑛𝑎𝑡ij𝑎𝑛i 3 − 𝑠𝑎𝑡𝑟𝑔𝑎 𝑦𝑜𝑧𝑎𝑚i𝑧^} 2 2 −1
^[ 0 −6 2 ^{] ={1 − 𝑠𝑎𝑡𝑟𝑔𝑎 3 − 𝑛i 𝑞𝑜′𝑠ℎ𝑎𝑚i𝑧 𝑣𝑎 𝑛𝑎𝑡ij𝑎𝑛i 3 −}
−2 −2 2
𝑠𝑎𝑡𝑟𝑔𝑎 𝑦𝑜𝑧𝑎𝑚i𝑧^} =
2 2 −1
^[0 −6 2 ^{] ={𝑑i𝑜𝑔𝑎𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑙𝑎𝑟𝑛i 𝑘𝑜′𝑝𝑎𝑦𝑡i𝑟i𝑙𝑎𝑑i}=-12}
0 0 1



𝐴
5-usul. A= ¹ (𝐴_ii · 𝐴_jj − 𝐴_ij · 𝐴_ji) formulada i=1 va j=2 bo’lsin. 𝐴₁₁₂₂, 𝐴₁₁,
iijj
𝐴₂₂, 𝐴₁₂, va 𝐴₂₁, - algebraik to’ldiruvchilarini hisoblaymiz:

𝐴₁₁₂₂ = 2 ; 𝐴₁₁ = [^{−4 1}]=-8+2=-6 𝐴₂₂ = [ ^{2 −1}] = 4 − 2 = 2

−2 2 −2 2

𝐴₁₂ = − [ ^{2 1}] = −⁽4 + 2⁾ = −6 𝐴₂₁ = − [ ^{2 −1}]=-(4-2)=-2

−2 2 −2 2
olingan natijalarni formulaga qo’yib, determinantning qiymatini topamiz:

A= ^1
𝐴iijj
(𝐴_ii · 𝐴_jj − 𝐴_ij · 𝐴_ji)=¹(-6· 2 − (−6)(−2)) = −12

2
Berilgan ikkinchi tartibli determinantlarni hisoblang.

_{1) [}−7 6



] 2) [^{10 −5}] 3).[^{10 −5}] 4) [^{√𝑎 + √𝑏 √𝑎 − √𝑏}]

5 −4
8 −8
9 −8
_√𝑎 − √𝑏 _√𝑎 + √𝑏

Berilgan uchinchi tartibli determinantni hisoblang.

	2	3	4		𝑎	1	𝑎		5	3	2		1	−1	4
1)	^[5	−2	1^]	2)	^[−1	𝑎	1^]	3)	^[−1	2	4^]	4)	^[3	−2	1 ^]
	1	2	3		𝑎	−1	𝑎		7	3	6		1	−1	−3

sin 𝑎	cos 𝑎	1	sin 𝑎	cos 𝑎	1	𝑐𝑡𝑔𝑏	𝑐𝑡𝑔𝑎	1
5) [^sin^𝑏	cos 𝑏	^1]	6) [^sin^𝑏	cos 𝑏	^0]	7) [𝑠i𝑛𝑎	𝑐𝑜𝑠𝑏	1]
sin 𝑦	cos 𝑦	1	sin 𝑦	cos 𝑦	0	𝑦	𝑦	1

Determinantning xossalaridan foydalanib, hisoblang.

1 2 3 4
− 1 − 1 − 1 − 1
1 2 0 − 3

_{1) [}−9 − 9 − 9 − 9_{] 2) [}
4 3 2 1
−1 − 2 − 4 − 8
−1 − 3 − 9 − 27
] 3) [
3 1 0 4 _]
1 5 − 1 7

1 0 1 0
−1 − 4 − 16 − 64
−2 1 0 1

_𝖥3 − 1 2 − 1 1_1
I5 1 − 2 1 2_I
⁴⁾ _I 9 − 1 1 3 4 _I
I 3 0 6 − 1 3 I
^[ 5 2 3 − 2 1 ^]
_𝖥3 − 1 2 − 1 1_1
I 5 1 − 2 1 2 _I
⁵⁾ _I9 − 1 1 3 4_I
I3 − 1 2 − 1 1 I
^[ 5 2 3 − 2 1 ^]

Nazorat misollari 1.Kvadrat matritsalarga misollar.

−1 3 3
9 2 6 3

^{A=[1 −6] B=[}−1 2 4^{] С=[1 5 8 2]}

^{2 4} 8 7 1
3 2 1 4
1 0 3 0

Matritsalar mos ravishda 2-tartibli,3-tartibli,4-tartibli matritsalar deyiladi.