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Предмет Алгебра

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Bohodir Jalolov

Chiziqli tenglamalar sistemasini yechish

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CHIZIQLI TENGLAMALAR SISITEMACINI YECHISH
Chiziqli tenglamalar sistemasini yechish
Reja
Chiziqli tenglamalar sistemasini solve buytug’I orqali yechish.
Chiziqli tenglamalar sistemasini Gaus usuli yordamida yechish.
Chiziqli tenglamalar sistemasini Kramer usuli yordamida yechish.
Chiziqli tenglamalar sistemasini teskari matritsa usuli yordamida yechish.
Matritsaviy tenglamini yechish.
 
        Chiziqli tenglamalar sistemasini yechishning oddiy usuli solve buyrug’I orqali bajarilad. 
Buning uchun maple matematik paketiga kirib solve buyrug’ini yozib ()qavs ichiga {} 
qavs ochamiz va berilgan tenglamalar sistemasini vergul bilan ajratib kiritamizva {}qavs 
tashqarisiga vergul quyib yechimni qaysi nomalumga nisbatan yechishini kiritamiz kata 
[]qavs bilan va qatorni ; yopib inter buyrug’ini beramiz va javobni qabul qilamiz.
        Gaus usuli bilan chiziqli tenglamalar sistemasini yechish uchun kutub honadan 
foydalanishga to’g’ri keladi bizga kerakli kutubhonani quyidagicha chaqiramiz 
>with(LinearAlgebra): va sistemaning asosiy matritsasini quyidagicha kiritamiz
A := <<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<-4,3,2,1>>;
Yordamch matritsasini esa > b := <5,1,1,-5>; ko’rinishda kiritamiz va Gaus usilidda 
ishlash uchun quyidagi buyruqlarni ketma ket kiritamiz   GaussianElimination(A); > 
GaussianElimination(A,'method'='FractionFree'); ReducedRowEchelonForm(<A|b>); 
 Natija toyyor.
                  Kramer usulida ishlash uchun ha with(LinearAlgebra): kutub honasidan 
foydalanamiz assos matritsasini yuqoridagidek kiritaniz va  Determinant(y)  ushbu 
buyruq orqali determinantini hisoblaymiz  yordamchi matritsani ketma ket ustunlar 
o’rniga almashtirib ulRNING HAM determinantini hisoblab olamiz va yakuniy natijani 
mos ravishda 
> x1:=Determinant(y1)/Determinant(y);
> x2:=Determinant(y2)/Determinant(y);
> x3:=Determinant(y3)/Determinant(y);
> x4:=Determinant(y4)/Determinant(y);
 Yechimlarni qabul qilib olamiz
        Teskari matritsa usulida yechish uchun ham yuqoridagi kutubhonadan foydalanamiz 
va asosmatritsasini topamiz so’ngunga teskari matritsasini topamiz  u quyidagicha  topiladi A^(-1); va topilgan teskari matritsaga yordamchi matritsani ko’paytiramiz  A^(-
1).B (.) buyerda matritsa ko’paytmasi amali chiqgan natija bizning Sistema yechimi.
 Mavzuga doir misollar 
Tenglamalar sistemasini Oddiy usul bilan yechish
> solve({x+2*y+3*z-4*b=5,2*x+y+2*z+3*b=1,3*x+2*y+z+2*b=1,4*x+3*y+2*z+b=-5},
[x,y,z,b]);[	]	[	]	,	,	,	x	-50	y	62	z	-3	b	15
 
Tenglamalar sistemasini Gauss metodi orqali yechish:
> with(LinearAlgebra):
A := <<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<-4,3,2,1>>;
 := A 







 







1 2 3 -4
2 1 2 3
3 2 1 2
4 3 2 1
> b := <5,1,1,-5>;
 := b 







 







5
1
1
-5
> GaussianElimination(A);













 












1 2 3 -4
0 -3 -4 11
0 0 -8
3 -2
3
0 0 0 -1
2 > GaussianElimination(A,'method'='FractionFree');








 







1 2 3 -4
0 -3 -4 11
0 0 8 2
0 0 0 -4
> ReducedRowEchelonForm(<A|b>);








 







1 0 0 0 -50
0 1 0 0 62
0 0 1 0 -3
0 0 0 1 15
 
Tenglamalar sistemasini Kramel usulida yechish:
> with(Student[LinearAlgebra]):
> y:= <<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<-4,3,2,1>>;
> y1 := < <5,1,1,-5>|<2,1,2,3>|<3,2,1,2>|<- 4,3,2,1>>;
 := y1 







 







5 2 3 -4
1 1 2 3
1 2 1 2
-5 3 2 1
> y2:= <<1,2,3,4>|<5,1,1,-5>|<3,2,1,2>|<-4,3,2,1>>;
 := y2 







 







1 5 3 -4
2 1 2 3
3 1 1 2
4 -5 2 1
> y3:= <<1,2,3,4>|<2,1,2,3>|<5,1,1,-5>|<-4,3,2,1>>; := y	


	


	
1	2	3	-4	
2	1	2	3	
3	2	1	2	
4	3	2	1  := y3 







 







1 2 5 -4
2 1 1 3
3 2 1 2
4 3 -5 1
> y4 := <<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<5,1,1,-5>>;
 := y4 



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 




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

1 2 3 5
2 1 2 1
3 2 1 1
4 3 2 -5
> x1:=Determinant(y1)/Determinant(y);	x1	-50
> x2:=Determinant(y2)/Determinant(y);	
	x2	62
> x3:=Determinant(y3)/Determinant(y);	
	x3	-3
> x4:=Determinant(y4)/Determinant(y);
Tenglamalarsistemasini Teskari matritsa yordamida echish
> with(Student[LinearAlgebra]):
> A:= <<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<-4,3,2,1>>;
 := A 







 







1 2 3 -4
2 1 2 3
3 2 1 2
4 3 2 1
> B:=<<5,1,1,-5>>;
 := B 







 







5
1
1
-5
> V:=A^(-1);  := V 

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-2 1
2 -8 13
2
5
2 -1 21
2 -8
0 1
2 -1 1
2
1
2 0 5
2 -2
> X:=V.B;
 := X 







 







-50
62
-3
15
 
Tenglamalar sistemasi
Oddiy usul bilan yechish
> solve({3*x+2*y+z=5,2*x+3*y+z=1,2*x+y+3*z=11},[x,y,z]);[	]	[	]	,	,	x	2	y	-2	z	3
Tenglamalar sistemasini Gauss metodi orqali yechish:
> with(LinearAlgebra):
> A:=<<3,2,2>|<2,3,1>|<1,1,3>>;
 := A 




 




3 2 1
2 3 1
2 1 3 > b:=<<5,1,11>>; := b	


	



5
1
11
> GaussianElimination(A);	


	


	
3	2	1	
0	5
3	
1
3	
0	0	12
5
> GaussianElimination(A,'method'='FractionFree');





 




3 2 1
0 5 1
0 0 12
> ReducedRowEchelonForm(<A|b>);





 




1 0 0 2
0 1 0 -2
0 0 1 3
 
Tenglamalar sistemasini Kramel usulida yechish
> with(Student [LinearAlgebra]):
> A:=<<3,2,2>|<2,3,1>|<1,1,3>>;  := A 




 




3 2 1
2 3 1
2 1 3
> b:=<<5,1,11>>; := b	


	



5
1
11
> X:=<<5,1,11>|<2,3,1>|<1,1,3>>;
 := X 




 




5 2 1
1 3 1
11 1 3
> Y:=<<3,2,2>|<5,1,11>|<1,1,3>>;	
 := Y	


	


	
3	5	1	
2	1	1	
2	11	3
> Z:=<<3,2,2>|<2,3,1>|<5,1,11>>;
 := Z 




 




3 2 5
2 3 1
2 1 11
> x:=Determinant(X)/Determinant(A);	
 := x	2
> y:=Determinant(Y)/Determinant(A);	
 := y	-2
> z:=Determinant(Z)/Determinant(A);
 := z 3
  Tenglamalarsistemasini Teskari matritsa yordamida echish
> with(Student [LinearAlgebra]):
> A:=<<3,2,2>|<2,3,1>|<1,1,3>>;
 := A 




 




3 2 1
2 3 1
2 1 3
> B:=<<5,1,11>>; := B	


	



5
1
11
> V:=A^(-1);
 := V 












 












2
3 -5
12 -1
12
-1
3 7
12 -1
12
-1
3 1
12 5
12
> X:=V.B; > with(Student [LinearAlgebra]):
> A:=<<4,1,3>|<-2,1,-2>|<0,2,0>>;
 := A 




 




4 -2 0
1 1 2
3 -2 0
> B:=<<0,2,0>|<-2,4,-3>|<6,3,4>>;
 := B 




 




0 -2 6
2 4 3
0 -3 4
> C:=A^(-1); := C	


	


	
1	0	-1	
3
2	0	-2	
-5
4	
1
2	
3
2
> X:=C.B;
 := X 




 




0 1 2
0 3 1
1 0 0

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