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                  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);