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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)
Ilmiybaza.uz
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
Ilmiybaza.uz
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.
