MAT1120 Linear Algebra (PDF)




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MAT1120 - Linear Algebra
Teodor Spæren
September 2015

1

Oppgave 1

Er gitt matrisen



P =



1.0
0
0
0
0

0.7
0
0.3
0
0

0
0
0.5
0
0 0.65
0.5
0
0 0.35

0
0
0
0
1.0








Skal beregne P k for k ∈ {2, 3, 4, 40, 80} og angi hva sannsynligheten er for at
systemet g˚
ar fra tilstand s4 til s2 i løpet av henholdsvis 2, 3, 4, 40 og 80 tidsskritt.
Bruker følgende matlab script for ˚
a gjøre dette:
P = [1 0.7 0 0 0; 0 0 0.5 0 0; 0 0.3 0 0.65 0; 0 0 0.5 0 0; 0 0 0 0.35 1];
for k = [2 3 4 40 80]
% P^k
pk = P^k
% This is the probability
round(pk(2,4),5)
end
P k blir da



P2 = 




P4 = 


P 80



=


1.0
0
0
0
0

0.7
0.15
0
0.15
0

0.35
0
0.475
0
0.175

1.0
0
0
0
0

0.805
0.07125
0
0.07125
0.0525

1.0
0
0
0
0

0.9
0
0
0
0.1

0
0.325
0
0.325
0.35

0.5163
0
0.2256
0
0.2581

0.6667
0
0
0
0.3333

0
0
0
0
1.0

0.2275
0.1544
0
0.1544
0.4637

0.4333
0
0
0
0.5667

0
0
0
0
1.0







,




P3 = 


0
0
0
0
1.0






1




,



P 40



=


1.0
0
0
0
0

0.805
0
0.1425
0
0.0525

1.0
0
0
0
0

0.9
0
0
0
0.1

0.35
0.2375
0
0.2375
0.175

0.6667
0
0
0
0.3333

0.2275
0
0.3088
0
0.4637

0.4333
0
0
0
0.5667

0
0
0
0
1.0

0
0
0
0
1.0












og pk24
p224 = 0.3250,

2

p324 = 0,

p424 = 0.1544,

p40
24 = 0,

p80
24 = 0

Oppgave 2

For ˚
a finne basisen for Nul(P − I5 ), ved ˚
a omgjøre matrisen [(P − I5 ) 0] til
redusert trappeform. Dette gjøres med matlab med kommandoen rref([(Peye(5)) zeros(5,1)]), noe som gir:



[(P −I5 ) 0] ∼ 



0
0
0
0
0

0.7
−1.0
0.3
0
0

0
0.5
−1.0
0.5
0

0
0
0
0
0.65 0
−1.0 0
0.35 0

gjør vi om dette til parameter form, f˚
ar vi:

x1
is




 x2 =
x3 =


x4 =



x5
is

0
0
0
0
0





 
 
∼
 
 

0
0
0
0
0

1.0
0
0
0
0

0
1.0
0
0
0

0 0
0 0
1.0 0
0 0
0 0

0
0
0
0
0

free
0
0
0
free

Fra dette kan vi se at enhver basis for Nul(P − I5 ) kan skrives som
 x1 
 x1 + x5 


0




0
β=




 x0
5

x1 + x5

a er det enkelt ˚
a se at P ikke er regulær da det ikke finnes en entydig likevektsvektor. Det at P ikke er regulær kunne vi ha misstenkt fra kalkulasjonene v˚
are
i Oppgave en, da selv for høye verdier av k var ikke P k strengt positiv. Dette er
derimot ikke nokk for ˚
a konkludere med at P ikke er regulær, da vi m˚
atte ha
sjekket for alle k ∈ N.

2













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