Int. J.IndustrialMathematisVol. 3,No. 1(2011)9-15
Gauss Elimination Algorithm for Interval Matrix
M.Adabitabarrozja
,F.babakordi,M.Shahhosseini
Departmentofmathematis,QaemshahrBranh,IslamiAzadUniversity, Qaemshahr,Iran.
Reeived7Deember2010; revised21February 2011;aepted21Marh2011.
|||||||||||||||||||||||||||||||-Abstrat
Let [A℄ = [[a
ij ℄℄
nn
be an interval for [a
ij ℄ = [a
ij ;a
ij
℄ (i,j=1,2,...,n). In transformation a
matrix into uppertriangular matrix by Gauss method, entries under themain diameter
shouldbezerobyelementaryrowoperations. Inthispaperweonsiderzeroasaninterval;
and,wewouldapplyGausseliminationmethodonintervalmatrixbyarithmetioperations
on intervals and thedenitionof omparisonmethod interval dates,we woulduse Gauss
elimination methodon intervalmatrix.
Keywords: Intervallinearequation;Gausseliminationalgorithm.
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1 Introdution
The problems of interval equations solution are of perennial interest, beause of their
diret relevane to pratial modeling and optimization of real-world proesses inluding
nane[2,4℄, eonomy[1,3,8,11℄, andmehanis[5℄. Gausseliminationmethodisusedfor
transformationmatrixintouppertriangularform. Thisuppertriangularformanbeused
forsolvingsystem[9℄.Gunterworked on feasibilityresultforintervalGaussianelimination
[6℄.Jurgen presenteda method by whih thebreakdown of interval Gaussian elimination
ausedbydivisionofanintervalontainingzeroanbeavoidedforsomelassesofmatries
[7℄. DymovaandSevastjanovproposed"intervalextendedzero"methodforsolvinginterval
and fuzzy equationsand they applied itforGauss eliminationalgorithm [10℄. We simply
seeifwe usethemethod,oeÆient matrixwillnotbe transformedinto uppertriangular
form.Therefor, we proposed a new method forsolving the problem. The organization of
the paper is as follows: In Setion 2, we will remind some basi denitions of interval
mathematis, intervalextended zero method,and Gauss elimination algorithm. "Interval
extendedzero"methodwillbeusedforGausseliminationalgorithminsetion3. Examples
willbeprovidedin Setion4and nally, Setion5 isalloated to onlusion.
2 Preliminaries
Denition 2.1. We all [a℄=[a;a ℄ an interval number if aa, and we onsider
arith-meti operations on these two intervalsof [a℄;[b℄ as follows:
[a℄+[b℄=[a+b;a+b℄; (2.1)
k[a℄=
[ka ;ka ℄ k 0;
[ka ;ka ℄ k <0
(2.2)
[a℄ [b℄=[a℄+( [b℄)=[a b;a b℄; (2.3)
[a℄[b℄=[minfab;a b;ab;abg;maxfab;ab;a b;abg℄; (2.4)
[a℄[b℄=[a℄[ 1
b ;
1
b
℄; (2.5)
And, we denetheomparison of thesetwointervals suh as:
[a℄=[b℄ !a=b ; a=b (2.6)
[a℄<[b℄ !a<b (2.7)
[a℄[b℄ !ab (2.8)
[a℄[b℄ !a+a=b+b (2.9)
[a℄[b℄ !a+a<b+b (2.10)
2.1 interval Extended zero
In problems with interval entries, it is better to onsider zero as a symmetri interval
aroundzero intheform of [ y;y℄. Inwhihy0 and we show itas[0
y
℄=[ y;y℄.
2.1.1 Properties of interval extendedzero
Zero satises some of the properties in real number set,for the present we show in the
followingequalitiesthat thepropertiessatisfyinintervalextended zero, too:
1. [0
y ℄+[0
y 0
℄=[ y;y℄+[ y 0
;y 0
℄=[ (y+y 0
);(y+y 0
)℄=[0
y+y 0
℄
2. [0
y ℄[0
y 0
℄=[ y;y℄[ y 0
;y 0
℄=[ yy 0
;yy 0
℄ =[0
yy 0
℄
3. [a℄+[0
y
℄=[a;a℄+[ y;y℄=[a y;a+y℄[a℄
4.
[a℄[0
y
℄ =[a;a ℄[ y;y℄
= 8
>
>
>
<
>
>
>
:
[ a y;ay℄[0
a y
℄; [a℄>0
[a y; ay℄[0
a y
℄; [a℄<0
[ maxf a y;ayg;maxf a y;ayg℄[0 ℄; 02[a℄
2.1.2 Solving interval equation
Dymova and Sevastjanov proposed interval extended zero method for solving a linear
equation as [a℄[x℄ [b℄ = [0℄ where [a℄;[b℄;[x℄ are interval numbers [10℄, and 0 2= [a℄. In
themethod,theygettheintervalsolutionofthisequationas[x℄=[x;x ℄byusingarithmeti
operations,weobtain:
1 If[a℄>0,[b℄>0,thenx= b
a
; x= b+b
a a b
a 2
2 If[a℄>0,[b℄<0,thenx= b+b
a a b
a 2
; x= b
a
3 If[a℄<0,[b℄<0,thenx= b
a
; x= b+b
a a b
a 2
4 If[a℄<0,[b℄>0,thenx= b+b
a a b
a 2
; x= b
a
5 If[a℄>0,02b,thenx= b
a
; x= b+b
a b
a
6 If[a℄<0,02b,thenx= b
a
; x= b+b
a b
a
2.1.3 Gauss elimination algorithm
Elementaryrowoperations:
1. R
ij
: we move i-throw and j-throwforeah other.
2. R
i
(): we multiplyi-throwin.
3. R
ij
(): wemultiplyi-throwin,and add theresult to j-throw.
In Gauss elimination algorithm, all entries under the main diameter should be zero by
elementary row operations to make matrix,upper triangular.In order to this matrix is
onsideredasA=[a
ij ℄
nn
,where a
ij
2<and a
ii
6=0. Let we areink th iterate. At the
beginningof k th iteration,A (k 1)
asfollows:
A (k 1)
= 2
6
6
6
6
6
6
6
6
6
4 a
(0)
11 a
(0)
12
::: ::: ::: a (0)
1n
0 a
(1)
21
::: ::: ::: a (1)
2n
.
.
. .
.
. .
.
.
.
.
.
0 ::: 0 a (k 1)
kk
::: a (k 1)
kn
.
.
.
.
.
. .
.
.
.
.
.
0 ::: 0 a (k 1)
nk
::: a (k 1)
nn 3
7
7
7
7
7
7
7
7
7
5
We put m
ik =
a (k 1)
ik
a (k 1)
k k
,all entries under main diameter will be zero by R
ki (m
ik )(i =
k+1;:::;n;k =1;2;:::;n 1)andwe have:
a (k)
ij =a
(k 1)
ij
+m
ik a
(k 1)
3 Interval extension of Gauss elimination algorithm
Let[A℄=[[a
ij ℄℄
nn
isa matrixof nnwith theentries [a
ij ℄=[a
ij ;a
ij
℄ (i,j=1,2,...,n). In
intervalGauss eliminationmethod,we have:
[a ij ℄ (k) =[a ij ℄ (k 1) +[m ik ℄[a kj ℄ (3.13)
InGauss eliminationmethodtheintervalmatrixshouldbe transformedinto upper
trian-gularmatrix. We solve thefollowingequation to obtain[m
ik ℄= [a ik ℄ (k 1) [a k k ℄ (k 1) : [a kk ℄ (k 1) [m ik ℄+[a
ik ℄
(k 1)
=0 (3.14)
Therefore, by intervalextendedzero methodweget:
[m
ik ℄=[m
ik ;m
ik
℄ (3.15)
Wherewewillobtain[m
ik
℄ from Setion(2.1.2),
1. If[a
kk ℄
(k 1)
>0,[a
ik ℄
(k 1)
>0,then
m ik = a ik (k 1) +a ik (k 1) a k k (k 1) + a k k (k 1) a ik (k 1) a k k (k 1)2 ; m ik = a ik (k 1) a k k (k 1)
2. If[a
kk ℄
(k 1)
>0,[a
ik ℄
(k 1)
<0,then
m ik = a ik (k 1) a k k (k 1) ; m ik = a ik (k 1) +a ik (k 1) a k k (k 1) + a k k (k 1) a ik (k 1) a k k (k 1)2
3. If[a
kk ℄
(k 1)
<0,[a
ik ℄
(k 1)
>0,then
m ik = a ik (k 1) a k k (k 1) ; m ik = a ik (k 1) +a ik (k 1) a k k (k 1) + a k k (k 1) a ik (k 1) a k k (k 1)2
4. If[a
kk ℄
(k 1)
<0,[a
ik ℄
(k 1)
<0,then
m ik = a ik (k 1) +a ik (k 1) a k k (k 1) + a k k (k 1) a ik (k 1) a k k (k 1)2 ; m ik = a ik (k 1) a k k (k 1)
5. If[a
kk ℄
(k 1)
>0,02[a
6. If[a
kk ℄
(k 1)
<0,02[a
ik ℄ (k 1) , then m ik = a ik (k 1) a k k (k 1) ; m ik = a ik (k 1) +a ik (k 1) a k k (k 1) + a ik (k 1) a k k (k 1)
Proposition 3.1. By R
ki ([m
ik
℄);is obtained [a
ik ℄
(k+1)
= [0
ik
℄ for (i = k +1;:::;n;k =
1;:::;n 1).
Proof: In R
ki ([m
ik
℄),i-th row is multiplied in [m
ik
℄ and the result is added to k-th
row,therefore,we obtain[m
ik ℄, if[a
kk ℄
(k 1)
>0,[a
ik ℄ (k 1) >0Then [m ik ℄=[m
ik ;m ik ℄=[ a ik (k 1) +a ik (k 1) a kk (k 1) + a kk (k 1) a ik (k 1) a kk (k 1)2 ; a ik (k 1) a kk (k 1) ℄ (3.16) Hene, [a ik ℄ (k) =[a ik ℄ (k 1) +[m ik ℄[a kk ℄ (k 1) =[a ik (k 1) ;a ik (k 1) ℄ +[ a ik (k 1) +a ik (k 1) a k k (k 1) + a k k (k 1) a ik (k 1) a k k (k 1)2 ; a ik (k 1) a k k (k 1) ℄[a kk (k 1) ;a kk (k 1) ℄ =[a ik (k 1) ;a ik (k 1)
℄+[ a
ik (k 1) a ik (k 1) + a k k (k 1) a ik (k 1) a k k (k 1) ; a k k (k 1) a ik (k 1) a k k (k 1) ℄
=[ a
ik (k 1) + a k k (k 1) a ik (k 1) a k k (k 1) ; a ik (k 1) a k k (k 1) a k k (k 1) +a ik (k 1)
℄=[0
ik ℄
Otherases are similar.
In (n 1)-th iteration,we getan uppertriangularmatrix withintervalzeroas:
A (n 1) = 2 6 6 6 6 6 6 4 [a 11 ℄ (0) [a 12 ℄ (0)
::: [a
1n ℄ (0) [0 21 ℄ [a 22 ℄ (1)
::: [a
2n ℄ (1) . . . . . . . . . . . . . . . . . . . . . . . . [0 n1
℄ ::: [0
nn 1 ℄ [a nn ℄ (n 1) 3 7 7 7 7 7 7 5 Where [a ij ℄ (k) =[a ij ℄ (k 1) +[m ik ℄[a kj ℄ (k 1)
;(k=1;2;:::;n 1) (3.17)
Remark 3.1. If wedon't haverowhanging,then the determinant ofinterval matrix will
beas follows:
j[A℄j= n Y i=1 [a ii ℄ (i 1) (3.18)
This obvious that, [A℄ is nonsingular if and only if02= [a ℄ (i 1)
4 Numerial examples
Here,we illustrateourmethod.
Example 4.1. Let us onsider
[A℄=
[2,3℄ [5,6℄
[3,4℄ [1,2℄
If we use the proposed method for Gauss elimination algorithm we will get:
[m
21 ℄=[m
21 ;m
21 ℄=[
5
3
; 1℄ (4.19)
Then withR
12 ([m
21 ℄)
[A (1)
℄=
[2,3℄ [5,6℄
[-2,2℄ [-9,-3℄
=
[2,3℄ [5,6℄
[0
21
℄ [-9,-3℄
Example 4.2. Let us onsider [A℄ matrix as follows [8℄:
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
[0.1389,0.1396℄ [0.0804,0.0806℄ [0.0033,0.0036℄ [0.0001,0.0001℄ [0.0321,0.0327℄ [0.0052,0.0054℄
[0.1565,0.1571℄ [0.5043,0.5047℄ [0.5634,0.5643℄ [0.3401,0.3421℄ [0.2405,0.2411℄ [0.2642,0.2654℄
[0.0001,0.0002℄ [0.0004,0.0005℄ [0.0067,0.0069℄ [0.0013,0.0015℄ [0.0090,0.0090℄ [0.0145,0.0149℄
[0.0110,0.01130℄ [0.0178,0.0178℄ [0.0296,0.0298℄ [0.0140,0.0143℄ [0.1103,0.0111℄ [0.0413,0.0417℄
[0.0214,0.0216℄ [0.0749,0.0750℄ [0.0917,0.0917℄ [0.00490,0.0490℄ [0.0400,0.0402℄ [0.0454,0.0458℄
[0.0268,0.0271℄ [0.0284,0.0407℄ [0.0142,0.0144℄ [0.0358,0.0363℄ [0.1086,0.1090℄ [0.0981,0.0987℄ 3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
We will get[A (5)
℄ as follows:
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4
[0.1389,0.1396℄ [0.0804,0.0806℄ [0.0033,0.0036℄ [0.0001,0.0001℄ [0.0321,0.0327℄ [0.0052,0.0054℄
[-0.0014,0.0014℄ [0.4131,0.4146℄ [0.5593,0.5606℄ [0.3400,0.342℄ [[0.2035,0.2051℄ [0.2581,0.2596℄
[-0.0001,0.0001℄ [-0.0001,0.0001℄ [0.0061,0.0065℄ [0.001,0.0013℄ [0.0088,0.0089℄ [0.0142,0.0147℄
[-0.0003,0.0003℄ [-0.0003,0.0003℄ [-0.0015,0.0015℄ [0.0015,0.0033℄ [0.081,0.0845℄ [-0.0008,0.004℄
[-0.0003,0.0003℄ [-0.0005,0.0005℄ [-0.0013,0.0013℄ [-0.003,0.003℄ [0.0698,0.1524℄ [-0.013,0.0010℄
[-0.0004,0.0004℄ [-0.0127,0.0127℄ [0.0177,0.00177℄ [-0.0231,0.0231℄ [-0.776,0.776℄ [-0.051,0.153℄ 3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
5 Conlusion
In Gauss elimination method all of entries under main diameter with elementary row
operationsshould be zero inorder to transform matrix into uppertriangular form; but,
we simply see if we use the proposed method in[10℄ for Gauss elimination method, the
matrix will not be transformed into upper triangular form. Therefore, in order to solve
this problem we proposed to use interval extended zero method for obtaining [m
Referenes
[1℄ J.J.Bukley, Solvingfuzzyequations ineonomis and nane,Fuzzy Sets and
Sys-tems48 (1992) 289-296.
[2℄ J.J.Bukley, The fuzzy mathematis of naneTriangular , Fuzzy Sets and Systems
21(1987) 257-273.
[3℄ J.J.Bukley, Y. Qu, Solvingsystems of linear fuzzy equayions, Fuzzy Sets and
Sys-tems43 (1991) 33-43.
[4℄ M.Li. Calzi, Towards a general setting for the fuzzy Mathematis of nane, Fuzzy
Setsand Systems35 (1990) 56-280.
[5℄ Sh.Chen,X.W. Yang, Intervalniteelement methodforbeamstruturesFinite
Ele-ments, Analysis andDesign 34 (2000) 75-88.
[6℄ M. Gunter, A ontribution to the feasibility of theinterval Gaussianalgorithm,
Re-liableomput. 12 (2006) 79-98.
[7℄ G.Jurgen,IntervalGaussianeliminationwithprovittighening,AppliedMathematis
30(2009) 1761-1772.
[8℄ Q.X. Li, Si.F. Liu, The fondation of the grey matrix and the grey input-out put
aalysis,AppliedMathematial Modelling32 (2008) 267-291.
[9℄ A. Neumaier, Interval method for systems of equations, ambridgeuniversity press,
ambridge, Uk, 1990.
[10℄ P. Sevastjanov, L. Dymova, A new method forsolving intervaland fuzzy equations:
Linearase, InformationSienes 179 (2009) 925-937.
[11℄ C.C.Wu,N.B.Chang,Greyinput-outputanalysisanditsappliation
