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CHIZIQLI ALGEBRAIK TENGLAMALAR SISTEMASINI TAQRIBIY
YECHISH USULLARI
Tayanch so‘z va iboralar: Chiziqli algebraik tenglamalar sistemasi yechimini
topishning iteratsion usullari
![Ilmiybaza.uz
III.1. Chiziqli algebraik tenglamalar sistemasi yechimini topishning iteratsion
usullari
a)
, ,
1.. ,
ij
А
a
i j
n n
tartibli kvadrat matritsa,
1
1
,...,
,
,...,
,
n
n
x
x
x
b
b
b
n
o‘lchovli vektorlar bo‘lsa, Ax=b
chiziqli sistema Gauss usuli bilan quyidagicha echiladi:
To‘g‘ri yurish. Kengaytirilgan [A|b] matritsa uchburchak matritsaga
keltiriladi.
b
nn
n
n
n
n
b
a
a
a
b
a
a
a
b
a
a
a
...
. .
. . . . . . .
.
...
...
2
1
2
2
22
21
1
1
12
11
~
)
1(
1
1( )
1( )
2
1)
(
1
2
(1)
2
1( )
22
1)
(
1
1
(1)
1
1( )
12
...
0
. . . . . . . .
.
...
0
...
1
nn
nn
n
n
n
n
n
a
a
a
a
a
a
a
a
a
~
)
(
1
1)
(
1
1
1)
(
1
2)
(
1
2
(2)
2
(2)
22
1)
(
1
1
(1)
1
(1)
13
(1)
12
2)
(
1
(2)
(2)
3
2)
(
1
(2)
3
(1)
33
2)
(
1
2
(2)
2
(2)
23
1)
(
1
1
(1)
1
(1)
12
(1)
12
0 0 ... 0 1
0
0 0 1
0
. . . . . . . . .
.
. . . . . . . . .
.
...
1
0
...
1
... ~
~
...
0
0
. .
.
.
.
.
. .
.
. .
.
.
.
. . . .
.
...
0
0
...
... 1
0
...
1
n
nn
n
n
n
n
n
n
n
n
n
n
nn
nn
n
n
n
n
n
n
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Teskari yurish. No’malumlar ketma-ket topiladi:
( )
( )
( )
1
1
1
,
,
1,
2,...,1.
n
n
i
i
n
nn
i
in
ij
i
j i
x
a
x
a
a x
i
n
n
b) Determinantni hisoblash.](/media/image/chiziqli-algebraik-tenglamalar-sistemasini-taqribiy-yechish-usullari-chiziqli-algebraik-tenglamalar-sistemasi-yechimini-topishning-iteratsion-usullari1099page_2.png "Ilmiybaza.uz
III.1. Chiziqli algebraik tenglamalar sistemasi yechimini topishning iteratsion
usullari
a)
, ,
1.. ,
ij
А
a
i j
n n
tartibli kvadrat matritsa,
1
1
,...,
,
,...,
,
n
n
x
x
x
b
b
b
n
o‘lchovli vektorlar bo‘lsa, Ax=b
chiziqli sistema Gauss usuli bilan quyidagicha echiladi:
To‘g‘ri yurish. Kengaytirilgan [A|b] matritsa uchburchak matritsaga
keltiriladi.
b
nn
n
n
n
n
b
a
a
a
b
a
a
a
b
a
a
a
...
. .
. . . . . . .
.
...
...
2
1
2
2
22
21
1
1
12
11
~
)
1(
1
1( )
1( )
2
1)
(
1
2
(1)
2
1( )
22
1)
(
1
1
(1)
1
1( )
12
...
0
. . . . . . . .
.
...
0
...
1
nn
nn
n
n
n
n
n
a
a
a
a
a
a
a
a
a
~
)
(
1
1)
(
1
1
1)
(
1
2)
(
1
2
(2)
2
(2)
22
1)
(
1
1
(1)
1
(1)
13
(1)
12
2)
(
1
(2)
(2)
3
2)
(
1
(2)
3
(1)
33
2)
(
1
2
(2)
2
(2)
23
1)
(
1
1
(1)
1
(1)
12
(1)
12
0 0 ... 0 1
0
0 0 1
0
. . . . . . . . .
.
. . . . . . . . .
.
...
1
0
...
1
... ~
~
...
0
0
. .
.
.
.
.
. .
.
. .
.
.
.
. . . .
.
...
0
0
...
... 1
0
...
1
n
nn
n
n
n
n
n
n
n
n
n
n
nn
nn
n
n
n
n
n
n
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Teskari yurish. No’malumlar ketma-ket topiladi:
( )
( )
( )
1
1
1
,
,
1,
2,...,1.
n
n
i
i
n
nn
i
in
ij
i
j i
x
a
x
a
a x
i
n
n
b) Determinantni hisoblash.")
Ilmiybaza.uz
III.1. Chiziqli algebraik tenglamalar sistemasi yechimini topishning iteratsion
usullari
a)
, ,
1.. ,
ij
А
a
i j
n n
tartibli kvadrat matritsa,
1
1
,...,
,
,...,
,
n
n
x
x
x
b
b
b
n
o‘lchovli vektorlar bo‘lsa, Ax=b
chiziqli sistema Gauss usuli bilan quyidagicha echiladi:
To‘g‘ri yurish. Kengaytirilgan [A|b] matritsa uchburchak matritsaga
keltiriladi.
b
nn
n
n
n
n
b
a
a
a
b
a
a
a
b
a
a
a
...
. .
. . . . . . .
.
...
...
2
1
2
2
22
21
1
1
12
11
~
)
1(
1
1( )
1( )
2
1)
(
1
2
(1)
2
1( )
22
1)
(
1
1
(1)
1
1( )
12
...
0
. . . . . . . .
.
...
0
...
1
nn
nn
n
n
n
n
n
a
a
a
a
a
a
a
a
a
~
)
(
1
1)
(
1
1
1)
(
1
2)
(
1
2
(2)
2
(2)
22
1)
(
1
1
(1)
1
(1)
13
(1)
12
2)
(
1
(2)
(2)
3
2)
(
1
(2)
3
(1)
33
2)
(
1
2
(2)
2
(2)
23
1)
(
1
1
(1)
1
(1)
12
(1)
12
0 0 ... 0 1
0
0 0 1
0
. . . . . . . . .
.
. . . . . . . . .
.
...
1
0
...
1
... ~
~
...
0
0
. .
.
.
.
.
. .
.
. .
.
.
.
. . . .
.
...
0
0
...
... 1
0
...
1
n
nn
n
n
n
n
n
n
n
n
n
n
nn
nn
n
n
n
n
n
n
n
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Teskari yurish. No’malumlar ketma-ket topiladi:
( )
( )
( )
1
1
1
,
,
1,
2,...,1.
n
n
i
i
n
nn
i
in
ij
i
j i
x
a
x
a
a x
i
n
n
b) Determinantni hisoblash.
Ilmiybaza.uz
(1)
(1)
12
1
12
1
(1)
(1)
22
2
21
22
2
11
11
1
1
2
(1)
(1)
2
1
...
...
0
...
...
det( )
. . . . . . .
,
. . . . . . . .
. . . .
. .
...
0
...
n
n
n
n
n
n
n
n
n
nn
n
nn
a
a
a
a
a
a
a
a
a
a
D
A
a
a D
a
a
a
a
a
(1)
(1)
22
2
(1)
1
11
(1)
(1)
2
...
1
. . . . . . ,
...
n
ij
ij
n
ij
ij
ij
n
nn
a
a
a
a
D
a
a a
a
a
a
.
Bo‘lishda ishtirok etgan elementlar
(1)
(
1)
11
, 22
,...,
n
nn
a
a
a
bosh elementlar
deyiladi. Buning davom ettirib quyidagi ifodaga kelamiz:
1)
(
(1)
22
11
1
11
...
n
nn
n
n
a
a a
a D
D
,
ya’ni determinant Gauss usulida bosh elementlarning ko‘paytmasiga teng.
v) Progonka usuli
0
0
0 1
0
1
1
1
,
,
1..
1,
i
i
i
i
i
i
i
n
n
n
n
n
b x
c x
d a x
b x
c x
d
i
n
a x
b x
d
uchburchak sistema uchun progonka usuli quyidagidan iborat
1
1
0
0
,
,
0,...,
1,
0
i
i
i
i
i
i
i
i
i
i
i
i
C
d
a u
v
u
i
n
u
v
a v
b
a v
b
(to‘g‘ri yurish),
1
1
,
,
1,
2,...,0.
n
n
n
n
i
i
i
i
n
n
n
d
a u
x
x
v
x
u
i
n
n
a v
b
g) D()=det(A-E)=(-1)n(n-p1- n-1-…-pn)=0
algebraik tenglama A-matritsaning xarakteristik tenglamasi deyiladi. Algebraning
asosiy teoremasiga asosan, D()=0 tenglama n ta ildizga ega. Bu ildizlar A-
![Ilmiybaza.uz
matritsaning xos sonlari deyiladi. Agar 12….n bulsa 1 son maksimal xos son
deyiladi. Uni iteratsiya usuli bilan topish mumkin: Ixtiyoriy y=y(0) vektor olinib
y(k)=Ay(k-1), k=1,2,… vektorlar tuziladi. Ko‘rsatiladiki,
( )
1
(
1)
lim
,
1.... ,
k
i
k
k
i
y
i
n
y
bu erda y=[y1,….,yn]. Ya’ni
n
i
y
y
y
y
k
i
k
i
k
i
k
i
1,....,
,
1)
(
1
)
(
1)
(
1
1)
(
2
D()=0 tenglama ildizlari turli usullar bilan topiladi.
Fadeev usulida det(A),A-1 ,p1,…,pn sonlar quyidagicha topiladi:
A1=A, Sp(A1)=p1, B1=A1-p1E,
A2=AB1, Sp(A2)/2=p2, B2=A2-p2E,
An=ABn-1, Sp(An)/n=pn, Bn=An-pnE, A-1=Bn-1/pn ,det(A)=pn.
III.2. Bitta variantning yechilish tartibi
1). Gauss usuli.
A matritsa koeffitsientlari
b o‘ng
tomon
3
1
1
5
1
3
1
5
1
1
3
5
1-qadam
1
1/3
1/3
5/3
0
8/3
2/3
10/3
0
2/3
8/3
10/3
2-qadam
1
1/3
1/3
5/3](/media/image/chiziqli-algebraik-tenglamalar-sistemasini-taqribiy-yechish-usullari-chiziqli-algebraik-tenglamalar-sistemasi-yechimini-topishning-iteratsion-usullari1099page_4.png "Ilmiybaza.uz
matritsaning xos sonlari deyiladi. Agar 12….n bulsa 1 son maksimal xos son
deyiladi. Uni iteratsiya usuli bilan topish mumkin: Ixtiyoriy y=y(0) vektor olinib
y(k)=Ay(k-1), k=1,2,… vektorlar tuziladi. Ko‘rsatiladiki,
( )
1
(
1)
lim
,
1.... ,
k
i
k
k
i
y
i
n
y
bu erda y=[y1,….,yn]. Ya’ni
n
i
y
y
y
y
k
i
k
i
k
i
k
i
1,....,
,
1)
(
1
)
(
1)
(
1
1)
(
2
D()=0 tenglama ildizlari turli usullar bilan topiladi.
Fadeev usulida det(A),A-1 ,p1,…,pn sonlar quyidagicha topiladi:
A1=A, Sp(A1)=p1, B1=A1-p1E,
A2=AB1, Sp(A2)/2=p2, B2=A2-p2E,
An=ABn-1, Sp(An)/n=pn, Bn=An-pnE, A-1=Bn-1/pn ,det(A)=pn.
III.2. Bitta variantning yechilish tartibi
1). Gauss usuli.
A matritsa koeffitsientlari
b o‘ng
tomon
3
1
1
5
1
3
1
5
1
1
3
5
1-qadam
1
1/3
1/3
5/3
0
8/3
2/3
10/3
0
2/3
8/3
10/3
2-qadam
1
1/3
1/3
5/3")
Ilmiybaza.uz
matritsaning xos sonlari deyiladi. Agar 12….n bulsa 1 son maksimal xos son
deyiladi. Uni iteratsiya usuli bilan topish mumkin: Ixtiyoriy y=y(0) vektor olinib
y(k)=Ay(k-1), k=1,2,… vektorlar tuziladi. Ko‘rsatiladiki,
( )
1
(
1)
lim
,
1.... ,
k
i
k
k
i
y
i
n
y
bu erda y=[y1,….,yn]. Ya’ni
n
i
y
y
y
y
k
i
k
i
k
i
k
i
1,....,
,
1)
(
1
)
(
1)
(
1
1)
(
2
D()=0 tenglama ildizlari turli usullar bilan topiladi.
Fadeev usulida det(A),A-1 ,p1,…,pn sonlar quyidagicha topiladi:
A1=A, Sp(A1)=p1, B1=A1-p1E,
A2=AB1, Sp(A2)/2=p2, B2=A2-p2E,
An=ABn-1, Sp(An)/n=pn, Bn=An-pnE, A-1=Bn-1/pn ,det(A)=pn.
III.2. Bitta variantning yechilish tartibi
1). Gauss usuli.
A matritsa koeffitsientlari
b o‘ng
tomon
3
1
1
5
1
3
1
5
1
1
3
5
1-qadam
1
1/3
1/3
5/3
0
8/3
2/3
10/3
0
2/3
8/3
10/3
2-qadam
1
1/3
1/3
5/3
![Ilmiybaza.uz
0
1
1/4
5/4
0
1
4
5
3-qadam
1
1/3
1/3
5/3
0
1
1/4
5/4
0
0
15/4
15/4
4-qadam
1
1
1
1
1
1
Gauss usuli bilan echib ushbu javobni oldik: x=[x1,x2,x3]=[1,1,1].
2). Zeydel iteratsiya usuli yechib javob olamiz:
( )
(
1)
(
1)
1
2
3
(5
)/3
k
k
k
x
x
x
( )
( )
(
1)
2
1
3
(5
)/3
k
k
k
x
x
x
( )
( )
( )
3
1
2
(5
)/3
k
k
k
x
x
x
0 0
0
0
1 x 1= 1.666667
x 2= 1.111111
x 3= 0.740741
2 x 1= 1.049383
x 2= 1.069959
x 3= 0.960219
3 x 1= 0.989941
x 2= 1.016613
x 3= 0.997815
4 x 1= 0.995190
x 2= 1.002331
x 3= 1.000826
5 x 1= 0.998948
x 2= 1.
x 3= 1.000326
6 x 1= 0.999866
x 2= 0.999936
x 3= 1.000066
3). Determinantni hisoblaymiz: Det(A)=20.
3
1
1
1 1/3 1/3
1
1/3
1/3
( )
1
3
1
31
3
1
3 0
8/3
2/3
1
1
3
1
1
3
0
2/3
8/3
1
1/3
1/3
1
1/3
1/3
1
1/3
1/3
8
8
8
5
8
5
3*
0
1
1/ 4
3*
0
1
1/ 4
3* *
0
1
1/ 4
3* *
20
3
3
3
2
3
2
0
2/3
8/3
0
0
5/ 2
0
0
1
Det A
4). Teskari matritsani hisoblash ( 3.5 punktda ko‘rsatilgan)](/media/image/chiziqli-algebraik-tenglamalar-sistemasini-taqribiy-yechish-usullari-chiziqli-algebraik-tenglamalar-sistemasi-yechimini-topishning-iteratsion-usullari1099page_5.png "Ilmiybaza.uz
0
1
1/4
5/4
0
1
4
5
3-qadam
1
1/3
1/3
5/3
0
1
1/4
5/4
0
0
15/4
15/4
4-qadam
1
1
1
1
1
1
Gauss usuli bilan echib ushbu javobni oldik: x=[x1,x2,x3]=[1,1,1].
2). Zeydel iteratsiya usuli yechib javob olamiz:
( )
(
1)
(
1)
1
2
3
(5
)/3
k
k
k
x
x
x
( )
( )
(
1)
2
1
3
(5
)/3
k
k
k
x
x
x
( )
( )
( )
3
1
2
(5
)/3
k
k
k
x
x
x
0 0
0
0
1 x 1= 1.666667
x 2= 1.111111
x 3= 0.740741
2 x 1= 1.049383
x 2= 1.069959
x 3= 0.960219
3 x 1= 0.989941
x 2= 1.016613
x 3= 0.997815
4 x 1= 0.995190
x 2= 1.002331
x 3= 1.000826
5 x 1= 0.998948
x 2= 1.
x 3= 1.000326
6 x 1= 0.999866
x 2= 0.999936
x 3= 1.000066
3). Determinantni hisoblaymiz: Det(A)=20.
3
1
1
1 1/3 1/3
1
1/3
1/3
( )
1
3
1
31
3
1
3 0
8/3
2/3
1
1
3
1
1
3
0
2/3
8/3
1
1/3
1/3
1
1/3
1/3
1
1/3
1/3
8
8
8
5
8
5
3*
0
1
1/ 4
3*
0
1
1/ 4
3* *
0
1
1/ 4
3* *
20
3
3
3
2
3
2
0
2/3
8/3
0
0
5/ 2
0
0
1
Det A
4). Teskari matritsani hisoblash ( 3.5 punktda ko‘rsatilgan)")
Ilmiybaza.uz
0
1
1/4
5/4
0
1
4
5
3-qadam
1
1/3
1/3
5/3
0
1
1/4
5/4
0
0
15/4
15/4
4-qadam
1
1
1
1
1
1
Gauss usuli bilan echib ushbu javobni oldik: x=[x1,x2,x3]=[1,1,1].
2). Zeydel iteratsiya usuli yechib javob olamiz:
( )
(
1)
(
1)
1
2
3
(5
)/3
k
k
k
x
x
x
( )
( )
(
1)
2
1
3
(5
)/3
k
k
k
x
x
x
( )
( )
( )
3
1
2
(5
)/3
k
k
k
x
x
x
0 0
0
0
1 x 1= 1.666667
x 2= 1.111111
x 3= 0.740741
2 x 1= 1.049383
x 2= 1.069959
x 3= 0.960219
3 x 1= 0.989941
x 2= 1.016613
x 3= 0.997815
4 x 1= 0.995190
x 2= 1.002331
x 3= 1.000826
5 x 1= 0.998948
x 2= 1.
x 3= 1.000326
6 x 1= 0.999866
x 2= 0.999936
x 3= 1.000066
3). Determinantni hisoblaymiz: Det(A)=20.
3
1
1
1 1/3 1/3
1
1/3
1/3
( )
1
3
1
31
3
1
3 0
8/3
2/3
1
1
3
1
1
3
0
2/3
8/3
1
1/3
1/3
1
1/3
1/3
1
1/3
1/3
8
8
8
5
8
5
3*
0
1
1/ 4
3*
0
1
1/ 4
3* *
0
1
1/ 4
3* *
20
3
3
3
2
3
2
0
2/3
8/3
0
0
5/ 2
0
0
1
Det A
4). Teskari matritsani hisoblash ( 3.5 punktda ko‘rsatilgan)
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1
0.4
0.1
0.1
0.1
0.4
0.1
0.1
0.1
0.4
A
III.3. Masalaning Mathcad dasturida yechilishi.
4 ta usulda yechish mumkin.
a) Teskari matritsa yordamida yechish
A
3
1
1
1
3
1
1
1
3
b
5
5
5
x
A 1
b
x
1
1
1
b) lsolve(A,b) protsedurasi yordamida yechish
s
lsolve A b
(
)
s
1
1
1
v) Given..Find bloki yordamida yechish
A
3
1
1
1
3
1
1
1
3
b
5
5
5
x
0
0
0
Given
A
x
b
x
Find x
( )
x
1
1
1
g) Teskari matritsa va determinantni hisoblash
A 1
0.4
0.1
0.1
0.1
0.4
0.1
0.1
0.1
0.4
d
A
d
20
d) Jordan –Gauss usuli (rref(A|b) bilan echish. Avvalo, augmetnt(A,b) komanda
yordamida kengaytirilgan matritsa B:=(A|b) tuziladi, so‘ng rref(A|b) komanda
(Jordan –Gauss usuli) V kengaytirilgan matritsaga qo‘llanilib sistema birlik
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matritsaga keltiriladi, so‘ng hosil bo‘lgan matritsadan oxirgi ustun (javob) A-
reduced komanda bilan kesib olinadi.
ORIGIN
1
A
3
1
1
1
3
1
1
1
3
b
5
5
5
B
augment A b
(
)
B
3
1
1
1
3
1
1
1
3
5
5
5
rref B
( )
1
0
0
0
1
0
0
0
1
1
1
1
x
rref B
( )
4
x
1
1
1
Mathcad dasturining qulay, soddaligi ko‘rinib turibdi.
III.4. Oddiy iteratsiya usulining Mathcad da tashkil etishni ko‘ramiz.
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×ÀÒÑ ó÷óí îääèé èòåðàöèÿ óñóëè
A
3
1
1
1
3
1
1
1
3
E
1
0
0
0
1
0
0
0
1
x
0
0
0
b
5
5
5
0.05
B
E
A
C
b
k
0 50
x
k 1
C
B x
k
x
20
0.997
0.997
0.997
x
30
1
1
1
x
10
0.9437
0.9437
0.9437
x
0
1
2
3
4
5
6
7
8
9
0
1
2
0
0.25
0.4375
0.5781
0.6836
0.7627
0.822
0.8665
0.8999
0.9249
0
0.25
0.4375
0.5781
0.6836
0.7627
0.822
0.8665
0.8999
0.9249
0
0.25
0.4375
0.5781
0.6836
0.7627
0.822
0.8665
0.8999
0.9249
III.5. Xos sonlar va xos vektorlarni Mathcad da topamiz.
A
3
1
1
1
3
1
1
1
3
r
eigenvals A
(
)
s
eigenvecs A
(
)
r
2
2
5
s
0.719
0.025
0.694
0.387
0.816
0.43
0.577
0.577
0.577
Eng katta xos soni iteratsiya usuli bilan topishni tashkil etamiz;
k
k
k
x
x
1
max
lim
,
.
1,0 ,...,
,
(0)
( 1)
берилган
x
A k
x k
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A
3
1
1
1
3
1
1
1
3
x
0
1
0
0
n
10
f x y
(
)
max
y3
x3
y1
x1
y2
x2
k
0 n
x
k 1
A
x
k
x
9
6.514
105
6.509
105
6.509
105
x
10
3.256
106
3.255
106
3.255
106
x
11
1.628
107
1.628
107
1.628
107
y
x
11
x
x
10
y0
x0
4.999
y1
x1
5
y2
x2
5
Demak,
max 5
, natija oldingi hisoblashlarni takrorlayapti.
Individual topshiriqlar.
Quyidagi sistema Gauss, Zeydelь, Fadeev usullari bilan echilsin va determinant,
teskari matritsa hisoblansin. Xos sonlar va xos vektorlar topilsin.
1).
8.30
2.62
4.10
1.90
16.92
3.92
8.45
7.78
2.46
22.61
,
,
0.2 ,
0,..., .
3.77
7.21
8.04
2.28
21.3
2.21
3.65
1.69
6.99
14.54
A
b
k k
n
,
2).
1
1
1
3
1
1
1
3
,
4,5,6,....,
.
1
1
1
3
1
1
1
3
k
k
k
k
A
k
b
k
k
k
k
Mavzu bo‘yicha savollar
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1. Iteratsiya usulining mohiyatini aytib bering.
2. Tenglama iteratsiya usulini qo‘llash uchun qanday qilib ko‘rinishga kelitiriladi.
3. Iteratsiya usulining yaqinlashish shartini ma’nosini aytib bering.
4. Iteratsiya usulining nazariy va amaliy xatoliklarining ma’nosini aytib bering.
