Post on 07-Aug-2018
8/20/2019 latihan metode numerik
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NAMA : DONI A.R PURBA
NIM : 122406108
GROUP : KC1 ‘12
1 SOAL PERTAMA
>> a=[1 ! "# 2 $ 8 1# 2 ! " $%
a =
1 ! "
2 $ 8 1
2 ! " $
>> a&2':(=a&2':()2*a&1':(
a =
1 ! "
0 1 )2 )!
2 ! " $
>> a&':(=a&':()2*a&1':(
a =
1 ! "
0 1 )2 )!
0 )1 )1 )11
8/20/2019 latihan metode numerik
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>> a&':(=a&':(*)1
a =
1 ! "
0 1 )2 )!
0 1 1 11
>> a&':(=a&':()a&2':(
a =
1 ! "
0 1 )2 )!
0 0 16
>> +=a&'4(,a&'(
+ =
!.
>> -=&a&2'4()+*a&2'((,a&2'2(
- =
!.666$
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>> = &a&1'4()+*a&1'()-*a&1'2((,a&1'1(
=
)4.666$
Buat soal dengan cara penyelesaian metode Gauss
>> a = [1 2 4 5; 2 9 3 7; 2 7 9 11]
a =
1 2 4 5
2 9 3 7
2 7 9 11
>> a(2,:) = a(2,:)-2*a(1,:)
a =
1 2 4 5
0 5 -5 -3
2 7 9 11
>> a(3,:) = a(3,:)-2*a(1,:)
a =
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1 2 4 5
0 5 -5 -3
0 3 1 1
>> a(3,:) = (5*a(3,:))-(3*a(2,:))
a =
1 2 4 5
0 5 -5 -3
0 0 20 14
>> z=a(3,4)/a(3,3)
z =
0.7000
>> y=(a(2,4)-z*a(2,3))/a(2,2)
y =
0.1000
>> x= (a(1,4)-z*a(1,3)-y*a(1,2))/a(1,1)
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x =
2
>> a = [1 3 5 7; 3 8 7 11; 2 5 9 18]
a =
1 3 5 7
3 8 7 11
2 5 9 18
>> a(2,:) = a(2,:)-3*a(1,:)
a =
1 3 5 7
0 -1 -8 -10
2 5 9 18
>> a(3,:) = a(3,:)-2*a(1,:)
a =
1 3 5 7
0 -1 -8 -10
0 -1 -1 4
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z =
2
>> y=(a(2,4)-z*a(2,3))/a(2,2)
y =
-6
>> x= (a(1,4)-z*a(1,3)-y*a(1,2))/a(1,1)
x =
15