Mapleda tenglamalar sistemasi va tenglamalarni yechish - Алгебра - Рефераты

Информация о документе

Цена	2000UZS
Размер	782.8KB
Покупки	0
Дата загрузки	07 Ноябрь 2026
Расширение	docx
Раздел	Рефераты
Предмет	Алгебра

Mapleda tenglamalar sistemasi va tenglamalarni yechish

Купить

Mavzu: Mapleda tenglamalar sistemasi va tenglamalarni yechish.
Reja:
1. Nazariy qism
2. Berilgan topshiriqning bajarilish qismi
3. Xulosa
4. Adabiyotlar
Quyidagi chiziqli tenglamlar sistemasini:
1) Gauss usulida eching.;
2) Kramer usulida eching.;
3) Matritsa tenglamasini teskari matritsa topish usulida eching.
Bunda tenglamalar sistemasi ko’rsatmada ko’rsatilganidek 2 ta
usul bilan hamda matritsali tenglama yechiladi.
Nazariy qism
Tenglamalar sistemasini yechish . Tenglamalar sistemasi ham xuddi
shunday solve({t1,t2,…},{x1,x2,…}) buyrug’i yordami bilan yechiladi, faqat endi
buyruq parametri sifatida birinchi figurali qavsda bir- biri bilan vergul bilan
ajratilgan tenglamalar, ikkinchi figurali qavsda esa noma’lum o’zgaruvchilar
ketma-ketligi yoziladi.
Masalan:
1)Tenglamalar sistemasini yeching.
x-y=1
x+y=3

>eq:={x-y=1,x+y=3};
eq := {x - y = 1, x + y = 3}
> s:=solve(eq,{x,y});
Enter tugmasini bosib natija:
s := {y = 1, x = 2}.
2)Tenglamalar sistemasini yeching
2x-2y=4
x+4y=6
> eq:={2*x-2*y=4,x+4*y=6};
eq := {x + 4 y = 6, 2 x - 2 y = 4}
> s:=solve(eq,{x,y});
Enter tugmasini bosib natija:
s := {y = 4/5, x = 14/5}
Mapleda tenglamalar sistemasini Gaus usulida yechib bo’lmaydi.
Kramer usulini Mapleda ishlash mumkin. Chunki Kramer usulida
matritsa va diterminatdan foydalanamiz. Kramer usuli quyidagicha
ishlanadi. Bizga quyidagi tenglamalar sistemasi berilgan bo’lsin.
KRAMER FORMULASINI QO’LLASH UCHUN ZARURIY SHART
YORDAMCHI DETERMINANTLARNI HISOLASH

TENGLAMALAR SISTEMASINI YECHISH UCHUN KRAMER FORMULASI
Mapleda matritsalar quyidagicha kiritiladi.
1 2 3 4
A= 5 6 7 8
9 8 6 5
>A:=<<1,5,9>|<2,6,8>|<3,7,6>|<4,8,5>>;
Ko’rinishida kiritiladi.
Diterminantni hisoblash uchun
>Determinant(A);
Buyrug’I kritiladi.
A matritsaning teskarisi deb A (-1)
ga aytiladi.
Bizga berilgan 3-misolda tenglama quyidagicha berilhgan.
A*X=B Bu yerda X no’malum matritsa. A va B matritsalar berilgan.
X=B/A bundan X=A (-1)
*B kelib chiqadi.
Berilgan topshiriqning bajarilish qismi
Berilgan tenglamalar sistemasini Gaus usulida yechamiz.

x
1 +2x
2 +3x
3 -4x
4 =5 (-2) (-3) (-4)
2x
1 +x
2 +2x
3 +3x
4 =1
3x
1 +2x
2 +x
3 +2x
4 =1
4x
1 +3x
2 +2x
3 +x
4 =-5
x
1 +2x
2 +3x
3 -4x
4 =5
-3x
2 -4x
3 +11x
4 =-9 (4) (5)
-4x
2 -8x
3 +14x
4 =-14 (-3)
-5x
2 -10x
3 +17x
4 =-25 (-3)
x
1 +2x
2 +3x
3 -4x
4 =5
-3x
2 -4x
3 +11x
4 =-9
8x
3 +2x
4 =6 (-5)
10x
3 +4x
4 =30 (4)
x
1 +2x
2 +3x
3 -4x
4 =5
-3x
2 -4x
3 +11x
4 =-9
8x
3 +2x
4 =6
6x
4 =90
Bundan x
4 =15, x
3 =-3, x
2 =62, x
1 =-50 ligi kelib chiqadi.
Berilgan tenglamalar sistemasini Maple dasturida yechamiz.
> a:={x1+2*x2+3*x3-4*x4=5, 2*x1+x2+2*x3+3*x4=1,
3*x1+2*x2+x3+2*x4=1, 4*x1+3*x2+2*x3+x4=-5};
a   x1 2 x2 3 x3 4 x4 5   2 x1 x2 2 x3 3 x4 1   3 x1 2 x2 x3 2 x4 1, , ,{ :=
  4 x1 3 x2 2 x3 x4 -5 }
> s:=solve(a,{x1,x2,x3,x4}); := s { } , , ,  x1 -50  x2 62  x3 -3  x4 15
> # Axtamov Javohir 203-guruh.
Endi Kramer usulida yechamiz.
KRAMER FORMULASINI QO’LLASH UCHUN ZARURIY SHART
> 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> Determinant(A);
-4
> # Axtamov Javohir 203-guruh.
Zaruriy shart bajarildi, Ya’ni - 4 ≠ 0 ga.
YORDAMCHI DETERMINANTLARNI HISOLASH
> x1:=<<5,1,1,-5>|<2,1,2,3>|<3,2,1,2>|<-4,3,2,1>>;
:= x1 







 







5 2 3 -4
1 1 2 3
1 2 1 2
-5 3 2 1
> x:=Determinant(x1);
:= x 200
> x2:=<<1,2,3,4>|<5,1,1,-5>|<3,2,1,2>|<-4,3,2,1>>;
:= x2 







 







1 5 3 -4
2 1 2 3
3 1 1 2
4 -5 2 1
> y:=Determinant(x2);
:= y -248
> x3:=<<1,2,3,4>|<2,1,2,3>|<5,1,1,-5>|<-4,3,2,1>>;
:= x3 







 







1 2 5 -4
2 1 1 3
3 2 1 2
4 3 -5 1
> z:=Determinant(x3);
:= z 12
> x4:=<<1,2,3,4>|<2,1,2,3>|<3,2,1,2>|<5,1,1,-5>>;
:= x4 







 







1 2 3 5
2 1 2 1
3 2 1 1
4 3 2 -5
> t:=Determinant(x4);
:= t -60
> # Axtamov Javohir 203-guruh.

Endi no’malumlarni topamiz.
> x1:=x/Determinant(A); := x1 -50
> x2:=y/Determinant(A);
:= x2 62
> x3:=z/Determinant(A);
:= x3 -3
> x4:=t/Determinant(A);
:= x4 15
> # Axtamov Javohir 203-guruh.
Berilgan tenglamalar sistemasini Gaus usulida yechamiz.
3x+2y+z=5 (-2) (-2)
2x+3y+z=1 (3)
2x+y+3z=11 (3)
3x+2y+z=5
5y+z=-7
-y+7z=23 (5)
3x+2y+z=5
5y+z=-7
36z=108
Bundan z=3, y=-2, x=2 ligi kelib chiqadi.
Berilgan tenglamalar sistemasini Maple dasturida yechamiz.
> restart;
> a:={3*x+2*y+z=5,2*x+3*y+z=1,2*x+y+3*z=11};
:= a { } , ,    2x y 3z 11    2x 3y z 1    3x 2y z 5
> solve(a,{x,y,z});
{ } , , x 2 y -2 z 3

> # Axtamov Javohir 203-guruh.
Endi Kramer usulida yechamiz.
KRAMER FORMULASINI QO’LLASH UCHUN ZARURIY SHART
> with(Student[LinearAlgebra]):
> A:=<<3,2,2>|<2,3,1>|<1,1,3>>;
:= A 




 




3 2 1
2 3 1
2 1 3
> Determinant(A);12
> # Axtamov Javohir 203-guruh.
Zaruriy shart bajarildi, Ya’ni 12 ≠ 0 ga.
YORDAMCHI DETERMINANTLARNI HISOLASH
> x:=<<5,1,11>|<2,3,1>|<1,1,3>>;
:= x 




 




5 2 1
1 3 1
11 1 3
> Determinant(x);
24
> y:=<<3,2,2>|<5,1,11>|<1,1,3>>;
:= y 




 




3 5 1
2 1 1
2 11 3
> Determinant(y);
-24
> z:=<<3,2,2>|<2,3,1>|<5,1,11>>;
:= z 




 




3 2 5
2 3 1
2 1 11
> Determinant(z);
36
> # Axtamov Javohir 203-guruh.

Endi no’malumlarni topamiz.
> x:=Determinant(x)/Determinant(A); := x 2
> y:=Determinant(y)/Determinant(A);
:= y -2
> z:=Determinant(z)/Determinant(A);
:= z 3
> # Axtamov Javohir 203-guruh.
A *X =B
Ko’rinishida berilgan matrisali tenglamani quyidagicha yechamiz.
X=B/A = A (-1)
*B
A (-1)
bu A matritsaning teskarisi.
Berilgan matritsali tenglamalani Maple dasturida yechamiz.
> restart;
> A:=<<4,1,3>|<-2,1,-2>|<0,2,0>>;
:= A 




 




4 -2 0
1 1 2
3 -2 0
> A^(-1);






1 0 -1
3
2 0 -2
-5
4
1
2
3
2
> B:=<<0,2,0>|<-2,4,-3>|<6,3,4>>;
:= B 




 




0 -2 6
2 4 3
0 -3 4
> X:=A^(-1).B;
:= X 




 




0 1 2
0 3 1
1 0 0
> # Axtamov Javohir 203-guruh.

Javob X matritsamiz:
:= X 




 




0 1 2
0 3 1
1 0 0
ga teng.

Foydalanilgan adabiyotlar
1.Maple 12 dasturining Help menyusi.
2.Olimov B.A , Masharipov M.P, Nuriddinova D. “ Matematik paket
“Maple” dasturida matematik masalalarni yechish”. Toshkent - 2016.
Foydalanilgan interset sayt
https://fr.maplesoft.com/products/Maple/index.aspx

Mapleda tenglamalar sistemasi va tenglamalarni yechish

Купить

Похожие документы
Ayirmali sxemalar, ularni qurish va differensial masalani approksimasiyalash. Dirixle masalasi. Chekli — ayirmali tenglamalar sistemasini iterasiya usuli bilan yechish (Libman o‘rtalash jarayoni)
Bernulli tenglamasi. puazeyl va darsi-veysbax formulalari. gidravlik zarba
Boshlang‘ich sinflarda matematikadan uy ishlari va o‘quvchilar bilimini reyting baholash va baho normalalri
Bir o‘lchamli nostatsionar issiqlik o‘tkazuvchanlik tenglamasini chekli ayirmalar usuli bilan yechish
Boshlang‘ich sinflarda matematika o‘qitish metodlari va ularning kilassifikatsiyasi