1-mavzu: Chiziqli algebraik tenglamalar sistemasini yechishning Gauss usuli

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chiziqli algebraik tenglamalar sistemasini yechishning gauss usuli

Bajarish.

1-mavzu: Chiziqli algebraik tenglamalar sistemasini

yechishning Gauss usuli

Nazariy va tadbiqiy matematikaning ko‘pgina masalalari birinchi

darajali chiziqli tenglamalar sistemasini yechishga olib kelinadi. Masalan,

funksiyaning n-ta nuqtada berilgan qiymatlari yordamida n-tartibli

ko‘phad bilan interpolyatsiyalash yoki funksiyani o‘rta kvadratlar usuli

yordamida yaqinlashtirish masalalari birinchi darajali chiziqli tenglamalar

sistemasini yechishga keltiriladi.

Birinchi darajali chiziqli tenglamalar sistemasini hosil qilishning

manbai uzluksiz funksional tenglamalarni chekli ayirmali tenglamalar

bilan yaqinlashtirishdir.

Birinchi darajali chiziqli tenglamalar sistemasini yechish asosan ikki

usulga, ya’ni aniq va iteratsion usullarga bo‘linadi.

Aniq usul deganda chekli miqdordagi arifmetik amallarni aniq bajarish

natijasida masalaning aniq yechimini topish tushuniladi.

Iteratsion usullarda chiziqli tenglamalar sistemasining yechimi ketma-

ket yaqinlashishlarning limiti sifatida topiladi.

Chiziqli tenglamalar sistemasini yechishning noma’lumlarni ketma-ket

yo‘qotish orqali aniqlash usuli, ya’ni Gauss usulini ko‘rib chiqamiz.

Bu usul bir necha hisoblash yo‘llariga ega. Shulardan biri Gaussning

kompleks yo‘lidir.

Ushbu sistema berilgan bo‘lsin











...

........

..........

,

...

...

11

n

n

n

nn

n

n

n

n

n

n

n

n

a

x

а

х

а

х

а

a

x

а

х

а

х

а

a

x

а

х

а

х

а

(1)

Faraz qilaylik, a

≠0 (etakchi element) bo‘lsin, aks holda

tenglamalarning o‘rinlarini almashtirib,

1

x

oldidagi koeffisienti noldan

farqli bo‘lgan tenglamani birinchi o‘ringa ko‘chiramiz.

Sistemadagi birinchi tenglamaning barcha koeffisientlarini a

bo‘lib,

)

(

)

(

)

(

...

+

n

n

n

b

x

b

x

b

х

(2)

ni hosil qilamiz, bu yerda

)

(

b

a

a

. . . ,

)

(

1

n

n

b

a

a

)

(

n

n

b

a

a

yoki qisqacha

(

)

(

≥

=

j

a

a

b

j

j

(2) tenglamadan foydalanib, (1) sistemaning qolgan tenglamalarida x

yo‘qotish mumkin. Buning uchun (2) tenglamani ketma-ket

a

21

, a

, … larga ko‘paytirib, mos ravishda sistemaning ikkinchi, uchinchi

va h.k. tenglamalaridan ayiramiz. Natijada, quyidagi sistema hosil bo‘ladi.











...

..........

...

)

(

)

(

)

(

)

(

)

(

)

(

22

n

n

n

nn

n

n

n

n

a

x

a

x

a

a

x

a

x

a

(3)

bu yerda

)

(

ij

a

koeffisientlar

)

1

(

)

(

j

i

ij

ij

b

a

a

a

−

(

)

≥

j

i

formula yordamida hisoblanadi.

Endi (3) sistema ustida ham shunga o‘xshash almashtirishlar

bajaramiz. Buning uchun (3) sistemadagi birinchi tenglamaning barcha

koeffisientlarini yetakchi element

)

(

≠

a

ga bo‘lib,

)

2

(

)

(

)

(

...

+

n

n

n

b

x

b

x

b

x

(4)

ni hosil qilamiz, bu yerda

)

(

)

(

)

(

2

a

a

b

j

j

)

(

≥

(4) tenglama yordamida (3) sistemaning keyingi tenglamalarida

yuqoridagidek x

2

ni yo‘qotib,











)

(

)

(

)

(

)

(

)

(

)

(

...

..........

...

n

n

n

nn

n

n

n

n

a

x

a

x

a

a

x

a

x

a

sistemaga kelamiz, bu yerda

)

(

)

(

)

(

)

(

j

i

ij

ij

b

a

a

a

−

(

)

≥

j

Noma’lumlarni yo‘qotish jarayoni davom ettirilib, bu jarayonni

m–qadamgacha bajarish mumkin deb faraz qilamiz va m – qadamda

quyidagi sistemaga ega bo‘lamiz.











...

......

..........

...

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

m

n

n

n

m

nn

m

m

m

n

m

n

m

n

m

n

m

m

m

m

m

m

n

m

n

m

mn

m

m

m

m

m

a

x

a

x

a

a

x

a

x

a

b

x

b

x

b

x

(5)

bu yerda

)

(

)

(

)

(

m

mm

m

mj

m

mj

a

a

b

)

(

)

(

)

(

)

(

m

mj

m

im

m

ij

m

ij

b

a

a

a

−

(

)

≥ m

j

i

Faraz qilaylik, m mumkin bo‘lgan oxirgi qadamning nomeri bo‘lsin.

Ikki hol bo‘lishi mumkin: m=n yoki m. Agar m=n uchburchak

matritsali va (1) sistemaga ekvivalent bo‘lgan quyidagi












=

=

+
+

+

=
+

+

+
+

+

+
+

)

(
1

,

)
2

(

1
,

2

)
2

(

2
3

)

2
(

23

2
)

1

(
1

,

1
)

1

(
1

3

)
1

(

13
2

)

1
(

12

1
,

..

..........

..........

..........

..........

..........

,
...

,

...

n

n

n

n

n

n

n

n

n

n

b

x

b

x

b

x

b

x

b

x

b

x

b

x

b

x
       (6)

sistemaga ega bo‘lamiz. Oxirgi sistemadan ketma-ket

1
1

...,

,
,

x

x

x

n

n

−
  larni

topish mumkin












−

−

=
−

=

=
+

−

−
−

+

−
−

−

.
...

..

..........

..........

..........

..........

)
1

(

,
1

2

)
1

(

1
,

1

1
)

1

(
,

1

)
1

(

1
,

1

1
)

(

1
,

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

x

b

x

b

x

x

b

b

x

b

x

                              (7)

(6) uchburchak sistemasining koeffisientlarini topish Gauss usulining

to‘g‘ri yurishi, (7) sistemadan yechimini topish Gauss usulining teskari

yurishi deyiladi.

Chiziqli tenglamalar sistemasini Gauss usuli yordamida yechish

algoritmi va dasturi

1-misol.

Gauss usuli bilan quyidagi sistema yechilsin.













=
+

+

+
=

+

−
+

=

−
−

+

=
−

+

−
,

2

,
2

3

2
4

,

4
2

2

3
,

5

4
2

3

2
4

3

2
1

4

3
2

1

4
3

2

1
4

3

2
1

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x



)
11

(

)

10
(

)

9
(

)

8
(

(8) tenglamadan x

1

ni topamiz

,
2
2

3

2
5

,

4
2

3

5
2

,

5
4

2

3
2

4

3
2

1

4
3

2

1
4

3

2
1

x

x

x

x

x

x

x

x

x

x

x

x

+
−

+

=
+

−

+
=

=

−
+

−



(12)
(12) tenglamani (9) tenglamadagi  x

1

  ni o‘rniga qo‘yamiz  va uni

ixchamlaymiz.

.
7

8

10
11

,

8
4

4

2
12

6

9
15

,

4
2

2

.
6

3

2
9

2

15
,

4

2
2

2

2
3

2

5
3

,

4
2

2

3
4

3

2
4

3

2
4

3

2
4

3

2
4

3

2
4

3

2
4

3

2
4

3

2
1

−

=
+

−

=
−

−

+
+

−

+
=

−

−
+

+

−
+

=

−
−

+











+
−

+

=
−

−

+

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

(12) tenglamani (10) tenglamadagi x

1

  ni o‘rniga qo‘yamiz va uni

ixchamlaymiz.

.
8

9

7

8
,

2

3
2

8

4
6

10

,
2

3

2
2

2

3
2

5

4
,

2

3
2

4

3
3

2

4
3

2

4
3

2

4
3

2

4
3

2

4
3

2

1
−

=

+
−

=

+
−

+

+
−

+

=
+

−

+










+

−

+
=

+

−
+

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

(12) tenglamani (11) tenglamadagi x

1

ni o‘rniga qo‘yamiz va uni

ixchamlaymiz.

Yuqoridagilardan quyidagi yangi tenglamalar sistemasini hosil qilamiz













−
=

+

−
=

+

−
−

=

+
−

=

−
+

−

)
15

(

.
1

6

5
)

14

(
,

8

9
7

8

)
13

(

,
7

8

10
11

,

5
4

2

3
2

4

2
4

3

2
4

3

2
4

3

2
1

x

x

x

x

x

x

x

x

x

x

x

x

(13) tenglamadan x

2

ni topamiz

,
11

8

11

10
11

7

,
8

10

7
11

,

7
8

10

11
4

3

2
4

3

2
4

3

2

x

x

x

x

x

x

x

x

x
−
+

−

=
−

+

−
=

−

=
+

−



(16)
.

1

6

5
,

4

2
2

2

4
2

3

5
,

2

2
2

3

2
5

,

2
4

2

4
3

2

4
3

2

4
3

2

4
3

2

4
3

2

1
−

=

+
=

+

+
+

+

−
+

=

+
+

+

+
−

+

=
+

+

+

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

(16) tenglamani (14) tenglamadagi
2

x   ni o‘rniga qo‘yamiz  va uni

ixchamlaymiz

,
32

35

3
,

88

99
77

64

80
56

,

8
9

7

11
64

11

80
11

56

,
8

9

7
11

8

11
10

11

7
8

,

8
9

7

8
4

3

4
3

4

3
4

3

4
3

4

3
4

3

4
3

2

−
=

+

−
=

+

−
−

+

−
−

=

+
−

−

+
−

−

=
+

−











−
+

−

−
=

+

−

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x


(15)
(16) tenglamani (15) tenglamadagi

2

x

  ni o‘rniga qo‘yamiz va uni

ixchamlaymiz

.
12

13

25
,

24

26
50

,

11
66

40

50
35

,

1
6

11

40
11

50

11
35

,

1
6

11

8
11

10

11
7

5

,
1

6

5
4

3

4
3

4

4
3

4

4
3

4

4
3

4

2
=

+

=
+

−

=
+

−

+
−

−

=
+

−

+
−

−

=
+











−

+
−

−

=
+

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

Yuqoridagilardan qo‘yidagi yangi tenglamalar sistemasini hosil qilamiz












=

+

−
=

+

−
=

+

−
=

−

+
−

)

18
(

.

12
13

25

)
17

(

,
32

35

3
,

7

8
10

11

,
5

4

2
3

2

4
3

4

3
4

3

2
4

3

2
1

x

x

x

x

x

x

x

x

x

x

x

(17) tenglamadan

3

x

ni topamiz

.
3

35

3

32
,

35

32
3

,

32
35

3

4
3

4

3
4

3

x

x

x

x

x

x

−
−

=

−
−

=

−
=

+



  (19)
(19) tenglamani (18) tenglamadagi

3

x

ni o‘rniga qo‘yamiz  va uni

ixchamlaymiz

[
]

.

1
,

836

836

,
36
13

875

800

,
12
13

3

875

3
800

,
12
13

3

35
3

32

25
,

12

13
25

4

4
4

4

4
4

4

4
4

3

−
=

=

−
=

+

−
−

=

+
−

−

=
+











−

−
=

+

x

x

x

x

x

x

x

x

x

x



(20)

(20) tenglamaning qiymatini (19) tenglamadagi

4

x ni o‘rniga qo‘yib

3

x  ni

topamiz.

[
]

.

1

,
1

3

35
3

32

3
35

3

32
3

4

3
=

=

+
−

=

−
−

=

x

x

x

                  (21)

(21) va (20) qiymatlarini (18) tenglamadagi

3

x va

4

x ni o‘rniga

qo‘yib

2

x ni topamiz.
,
1

11

11
11

8

11
10

11

7
11

8

11
10

11

7
4

3

2
=

=

+
+

−

=
−

+

−
=

x

x

x

                       (22)

[
]

.

1

2
=

x

(20),  (21) va(22) larni qiymatlarini  (12) tenglamadagi  x

2

,  x

3
va  x

4
lar  ni

o‘rniga qo‘yib x

1
ni topamiz.

[
]

.

1

,
1

3

4
3

5

,
1

5

,
2

2

1
1

2

3
2

5

2
2

3

2
5

1

4
3

2

1
=

=

−
=

−

+
=

−

−
⋅

+

=
+

−

+
=

x

x

x

x

x

Demak,  topilgan ildizlar

[
]

1

1
=

x

,
[

]

1
2

=

x

,
[

]

1
3

=

x

,
[

]

1
4

−

=

x

  berilgan

tenglamalar sistemasini to‘liq qanoatlantiradi.

Tenglamalar sistemasi qo‘lda yechilganda hisoblashlarni  1-jadvalda

ko‘rsatilgan Gaussning kompakt sxemasi bo‘yicha olib borish ma’quldir.

Soddalik uchun jadvalda to‘rtta no’malumli to‘rtta tenglamalar sistemasini

yechish sxemasi keltirilgan.

x

1

x

2

x

3

x

4

OZOD

HADLAR

∑
SXEMA

QISMLARI

a

11

a

21

a

31

a

41

…

1

a

12

a

22

a

32

a

42

…

b

12

(1)

a

13

a

23

a

33

a

43

…

b

13

(1)

a

14

a

24

a

34

a

44

…

b

14

(1)

a

15

a

25

a

35

a

45

…

b

15

(1)

a

16

a

26

a

36

a

46

…

b

16

(1)

A

a

22

(1)

a

32

(1)

a

42

(1)

…

a

23

(1)

a

32

(1)

a

43

(1)

…

a

24

(1)

a

34

(1)

a

44

(1)

…

a

25

(1)

a

35

(1)

a

45

(1)

…

a

25

(1)

a

35

(1)

a

45

(1)

…

A

1

1-jadval

1-jadvalda keltirilgan Gaussning kompakt sxemasi yordamida quyidagi

tenglamalar sistemasi yechilsin:

2-misol.












=

+

−
−

−

=
−

+

−
−

=

−
+

−

=
−

+

+
.

0

5
,

1

2
,

1

8
,

0

6
,

1

,
6

,

1
4

,

2
3

4

,
0

,

2
,

3

3
6

,

1
2

,

4
2

4

3
2

1

4
3

2

1
3

2

1
4

3

2
1

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

Sistemani yechish jarayoni 2-jadvalda keltirilgan.

2 - jadval

1

b

23

(2)

b

24

(2)

b

25

(2)

b

25

(2)

a

33

(2)

a

43

(2)

...

1

a

34

(2)

a

44

(2)

…

b

34

(3)

a

35

(2)

a

45

(2)

…

b

35

(3)

a

36

(2)

a

46

(2)

…

b

36

(3)

A

2

a

44

(3)

…

1

a

45

(3)

…

b

45

(4)

a

46

(3)

…

b

46

(4)

A

3

1

1

1

1

x

4

x

3

x

2

x

1

x

4

x

3

x

2

x

1

B

1

х

2

х

3

х

4

х

OZOD

HADL

AR

∑

SXEMA

QISMLA

RI

2

-0,4

1,6

1

…

1

4,2

3

-0,8

-2

…

2,1

1,6

-2,4

1

-1

…

0,8

-3

0

-1

1,5

…

-1,5

3,2

-1,6

-1

0

…

1,6

8

-1,4

-0,2

-0,5

…

4

A

3,84

4,16

-2,08

0,28

-0,60

-1,40

-0,96

3,56

0,2

6,6
1

А

      Shunday qilib, quyidagi

1

х

=1,00002 ,

2

х

=2,00005 ,

3

х

=3,00009 ,

4

х

=4,00013  taqribiy  yechimga  ega  bo‘ldik.  Sistemaning  aniq  yechimi

1

х

=1,
2

х

=2,

3

х

=3,
4

х

=4 ekanligiga bevosita ishonch hosil qilish mumkin.

Misol. Quyidagi ko‘rinishdagi chiziqli algebraik tenglamalar sistemasini

Gauss usuli yordamida yechish algoritmi va dasturini tuzing.

11
1

12

2

1
1

1

21
1

22

2
2

2

1
1

1

2
2

1

...

...
......

...

n

n

n

n

n

n

n

n

nn

n

nn

a x

a

x

a

x

a

a

x

a

x

a

x

a

a

x

a

x

a

x

a
+
+

+

+
+

+

=




+
+

+

=







+

+
+

=



4,1

…

1

1,8

…

-

0,5416

6

-3

…

-

0,15625

1,6

…

-0,25

4,5

…

0,05208

-

2,5333

1

-

4,0208

1

…

1

0,75

2,35937

…

-

0,29606

-4,6

-

2,62500

…

1,81581

-6,38331

-4,28644

…

2,51198

2

А

1,16897  4,67603

5,84500
3

А

1

1

1

1

4,00013

3,00009

2,00005

1,00002

5,00013

4,00009

3,00005

2,00002

В

Algoritmi:


begin

n

j

i

a
,

n

i

,
1

=

,

1

=

n

j

+

+

k

i

a

c
,
1

+

=

1
,

+

=

n

k

j

j

i

k

k

j

k

j

i

a

c

a

a

a
,
1

,

,
,

1

+
+

−

=

1
,

−

=

n

k

i

0

,
1

=

+ k

l

a

l=k

n

k
,
1

=

0

,
=

k

k

a

l=l+1

1
,

+

= n

k

p

p

k

a

d
,
=

p

l

kp

a

a

,
1

+

=

d

a

p

l
=
+ ,

1

1



S=0

n

n

n

n

n

a

a

x
,
1

,

+
=

n

k

i

,
1

+

=

i

i

k

x

a

S

S

,
+

=

1

,

1
−

= n

k

k

k

n

k

k

a

S

a

x
,
1

,

−
=

+

n

i

,
1

=

i

x

end

1

Dastur i:

11
1

12

2
1

1

21
1

22

2
2

2

1
1

2

2
...

...

......

...

n

n

n

n

n

n

nn

n

n

a

x

a

x

a

x

b

a

x

a

x

a

x

b

a

x

a

x

a

x

b
+
+

+

=




+
+

+

=







+

+
+

=



Program Gauss1;

label 1,2,3,4,5;

var a:array[1..10, 1..10] of real;

b,x:array[1..10] of real;

c,s:real; i,j,k,n:integer;

begin

  readln(n);

  for i:=1 to n do

  begin

  for j:=1 to n do

  read(a[i,j]);

  readln(b[i]);

  end;

  k:=1;

3: i:=k+1;

2: c:=a[i,k]/a[k,k];

  a[i,k]:=0;

  j:=k+1;

1: a[i,j]:=a[i,j]-c*a[k,j];

  if j
  b[i]:=b[i]-c*b[k];

  if i
  if k
  x[n]:=b[n]/a[n,n];

  i:=n-1;

5: j:=i+1;

  s:=0;

4: s:=s+a[i,j]*x[j];

   if j
  x[i]:=(b[i]-s)/a[i,i];

  if i>1 then begin i:=i-1; goto 5 end;

  for i:=1 to n do

  writeln(x[i]:4:2);

end.

11
1
12

2

1
1

1

21
1

22

2
2

2

1
1

1

2
2

1

...

...
......

...

n

n

n

n

n

n

n

n

nn

n

nn

a

x

a

x

a

x

a

a

x

a

x

a

x

a

a

x

a

x

a

x

a
+
+

+

+
+

+

=




+
+

+

=







+

+
+

=



program Gauss;

var a:array[1..10, 1..10] of real;

x:array[1..10] of real;

c,s,d:real; i,j,k,n,l,p:integer;

begin

  readln(n);

  for i:=1 to n do

  for j:=1 to n+1 do

readln(a[i,j]);

  for k:=1 to n do

  begin

  l:=k;

  while a[k,k]=0 do

  begin

   if a[l+1,k]=0 then else

  begin

  for p:=k to n+1 do7

  begin

  d:=a[k,p];

  a[k,p]:=a[l+1,p];

  a[l+1,p]:=d;

  end;

  break;

  end;

  l:=l+1;

  end;

  for i:=k to n-1 do

  begin

  c:=a[i+1,k];

  for j:=k to n+1 do

a[i+1,j]:=(a[k,j]/a[k,k])*c-a[i+1,j];

  end;

  end;

  x[n]:=a[n,n+1]/a[n,n];

  for k:=n-1 downto 1 do

  begin

s:=0;

for i:=k+1 to n do

s:=s+a[k,i]*x[i];

x[k]:=(a[k,n+1]-s)/a[k,k]

  end;

  for i:=1 to n do

  writeln(x[i]:4:2);

end.

2-masala.  Quyidagi chiziqli tenglamalar sistemasini yeching:

1
2

3

4
1

2

3
1

3

4
1

2

3
4

3

5
7

2

5
3

1

2
4

3

6
6

4

3
2

3

x

x

x

x

x

x

x

x

x

x

x

x

x

x

−
+

+

=




+
−

= −


− −

+

=




+
−

−

=


Bajarish.  1-masaladagidek, tenglamalar sistemasini

B

AX
=
ko`rinishda

yozib olamiz. Bu yerda A  –  noma`lumlar koeffisentlardan tashkil topgan

matritsa,  B–  ozod hadlardan tashkil topgan ustun (vektor), X–  noma`lumlar

ustuni (vektori).

3 -1 5 1

2 5 -3 0

-2 0 -4 3

6 4 -3 -2

A











=










,

7

1

B

 
=  

−

 

,
1
2

x

X

x

 

=  
 

Demak,

B

A

X
1
−

=

.
A  matritsani, ya`ni noma`lumlar koeffisentlarini A1:D4  maydonga,  B

vektorni, ya`ni ozod hadlarni F1:F4  maydonga  kiritamiz.  X  vektor uchun

H1:H4  maydonni belgilab

=МУМНОЖ(МОБР(A1:D4);F1:F4)  formulani

kiritamiz va Ctrl+Shift+Enter  tugmalarini birgalikda bosamiz. Natijada

H1:H4 maydonda izlanayotgan noma`lumlar hosil bo`ladi:

Document Outline

1-mavzu: Chiziqli algebraik tenglamalar sistemasini

sistemaga kelamiz, bu yerda

(12) tenglamani (11) tenglamadagi xR1 Rni o‘rniga qo‘yamiz va uni ixchamlaymiz.

Yuqoridagilardan quyidagi yangi tenglamalar sistemasini hosil qilamiz

(13) tenglamadan xR2 Rni topamiz

(16)

(16) tenglamani (15) tenglamadagi ni o‘rniga qo‘yamiz va uni ixchamlaymiz

Yuqoridagilardan qo‘yidagi yangi tenglamalar sistemasini hosil qilamiz

(17) tenglamadan ni topamiz

(19) tenglamani (18) tenglamadagi ni o‘rniga qo‘yamiz va uni ixchamlaymiz

(20) tenglamaning qiymatini (19) tenglamadagi ni o‘rniga qo‘yib ni topamiz.

Tenglamalar sistemasi qo‘lda yechilganda hisoblashlarni 1-jadvalda ko‘rsatilgan Gaussning kompakt sxemasi bo‘yicha olib borish ma’quldir.

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