CHIZIQLI TENGLAMALAR SISTEMACINI YECHISH
Reja: Kirish. 1.Chiziqli tenglamalar sistemasini Gauss usulida hisoblash 2.Chiziqli tenglamalar sistemasini Kramer usulida hisoblash 3.Matritsa tenglamasini teskari matritsa toppish usulida hisoblash Xulosa . Foydalanilgan adabiyotlar.
1. CHIZI Q LI TENGLAMALAR SISTEMACINI Y ECHISH Chiziqli tenglamalar sistemasini Gauss, Kramer va teskari matritsa usullari bilan y echimini topish masala la rini ko‘ramiz. 1.1-masala. Quyidagi uch noma’lumli chiziqli tenglamalar sistemasi ning yechimini: 1) Gauss usuli, 2) Kramer, 3) Matritsa usulida toping. 39 16 25 5 18 12 14 3 0 13 7 2 3 2 1 3 2 1 3 2 1 x x x x x x x x x (1) 1.1. Gauss usulida y echish. Yetakchi tenglama uchun birinchi tenglamani olamiz. Bu tenglamadan etakchi noma’lum uchun x 1 va a 11 ≠0 ni etakchi element uchun tanlaymiz. Birinchi tenglama dagi x 1 ning koeffitsenti a 11 ni 1 ga aylantirish uchun birinchi tenglamani ng barcha qo‘shiluvchilarini a 11 ≠0 ga b o‘ lamiz . Ҳosil bo‘lgan tenglamadan foydalanib ikkinchi va uchinchi tenglmalardan x 1 nomahlumni yo‘qotish yoki uning koeffitsentini nolg‘ga aylantirish uchun etakchi tenglamani –3 ga ko‘paytirib 2- tenglamaga qo‘shamiz, so‘ngra etakchi tenglamani –5 ko‘paytirib 3- teglamaga qo‘shamiz. Natijada quyidagicha sistemaga kelamiz: 39 2 33 2 15 18 2 15 2 7 0 13 7 2 3 2 3 2 3 2 1 x x x x x x x Bu tenglamlar sistemasida etakchi tenglama uchu 2- tenglamani olamiz. Unda 7/2 koeffitsient l i x 2 7 3 7 3 18 2 15 2 7 0 13 7 2 332 321 x x x x x x bu sistemaning 3- tenglamasidan x 3 nomahlumni topamiz. x 3 asosida 2- tenglamadan x 2 ni topamiz. x 3 , x 2 lar asosida 1- tenglamadan x 1 ni topamiz. x 3 = - 1 x 2 = 3 )2 15 18(7 2 ) 2 15 18(7 2 3 x x 1 = 4 ))1 ( 13 3 7( 2 1 ) 13 7( 2 1 3 2 x x Demak sistema echimi: x 1 =-4, x 2 =3, x 3 = -1
Maple12 dasturida masalani echish. Uch noma’lumli chiziqli tenglamalar sistemasini oddiy va Gauss usulida echish( 1 . 1 - masala). 1 2 3 1 2 3 1 2 32 7 13 0, 3 14 12 18, 5 25 16 39, ő ő ő ő ő ő ő ő ő Maple7 dasturida masalalarni echishdagi amallarni bajarish uchun ishchi oynada > belgidan so‘ng kerakli buyruqni yozib Enter tugmasini bosish kerak. 1. Oddiy usulida echish ( Gauss.mw ). > solve( {2*x + 7*y + 13*z = 0, 3*x + 14*y + 12*z =18, 5*x + 25*y +16*z =39}, [x, y, z]); [x=-4, y=3, z=-1] 2. Gauss usulida uch noma’lumli chiziqli tenglamalar sistemasini echish . > with(LinearAlgebra): A := <<2,3,5>|<7,14,25>|<13,12,16>>; 2 7 13 : 3 14 12 5 25 16 A > B := <0,18,39>; 0 : 18 39 b > GaussianElimination(A); 2 7 13 7 15 0 3 2 3 0 0 7 > GaussianElimination(A,'method'='FractionFree'); 2 7 13 0 7 15 0 0 3 >ReducedRowEchelonForm(<A|b>); 1 0 0 4 0 1 0 3 0 0 1 1
3. To‘rt noma’lumli chiziqli tenglamalar sistemasini Maple12 dasturida echish 1) Oddiy usulida echish > sys:=({1*x1-5*x2-1*x3+3*x4=-5,2*x1+3*x2+1*x3-1*x4=4, 3*x1-2*x2+3*x3+4*x4=- 1,5*x1+3*x2+2*x3+2*x4=0}): > solve(sys,{x1,x2,x3,x4});{x4 = K 3,x2 = K 1,x1 = 1,x3 = 2} 2) Gauss usulida echish 0.= 2x+ 2x+ 3x+ 5x -1,= 4x+ 3x+ 2x- 3x 4,= x- x3+ 3x+ 2x -5,= 3x+ x- 5x- x 4 3 2 1 4 3 2 1 4 2 1 4 3 2 1 > with(LinearAlgebra): A := <<1,2,3,5>|<-5,3,-2,3>|<-1,1,3,2>|<3,-1,4,2>>; 1 5 1 3 2 3 1 1 : 3 2 3 4 5 3 2 2A > b := <-5,4,-1,0>; 5 4 : 1 0 B > GaussianElimination(A,'method'='FractionFree'); 39/ 67 0 0 0 2 3 0 0 7 3 13 0 3 1 5 1 > ReducedRowEchelonForm(<A|b>); 3 0 0 0 2 1 0 0 1 0 1 0 1 0 0 1
1. 2. Kramer q oidasi yordamida echish. Berilgan tenglamalar sistema nomahlumlarning koeffitsientlari va ozod hadalari yordamida determinantlarni tuzamiz va ularni hisoblashning uchburchak yoki Sarrus usullaridan foydalanamiz. Biz (1) tenglamlar sistemasining determinantlarini tuzib, uchburchak usulida hisolab uni son qiymatlarni topamiz. 16255 21143 1375 3 336 600 910 420 975 448 16 7 3 25 2 12 13 14 5 5 12 7 13 25 3 26 14 2 12 2016 0 7098 5850 3276 0 16 25 39 12 14 18 13 7 0 1 x 99360117015210576 163916 12183 1302 2 x x 3 = 3 819 900 0 630 0 1092 39 25 5 18 14 3 0 7 2 Maple12 dasturida masalani echish. > with(Student[LinearAlgebra]): > d := <<2,3,5>|<7,14,25>|<13,12,16>>; > d:=Determinant(d); 39/ 67 0 0 0 2 3 0 0 7 3 13 0 3 1 5 1 > ReducedRowEchelonForm(<A|b>); 3 0 0 0 2 1 0 0 1 0 1 0 1 0 0 1