TUGAS

METODE NUMERIK

DOSEN :

HERU DIBYO LAKSONO, MT.

Oleh :

ILHAM FAJRI

1210953032

JURUSAN TEKNIK ELEKTRO

FAKULTAS TEKNIK

UNIVERSITAS ANDALAS

PADANG

2014

Soal :

2.51 X1 + 1.48 X2 + 4.53 X3 = 0.05

1.48 X1 + 0.93 X2 - 1.30 X3 = 0.64 (BP 32 x 2 : 100 = 0.64)

2.68 X1 + 3.04 X2 – 1.48 X3 = -0.53

Ditanya : Nilai X1, X2, X3 = . . . ?

Penyelesaian :

[2.51 1. 48 4. 531.48 0.93 −1.302.68 3. 04 −1.48] [X1

X2

X3]=[ 0.05

0.64−0.53 ]

Baris 1 :

2.51 : 2.51 = 1

1.48 : 2.51 = 0.59

4.53 : 2.51 = 1.80

0.05 : 2.51 = 0.019

[ 1 0.59 1. 801.48 0.93 −1.302.68 3.04 −1.48 ][ 0.019

0.64−0. 53]

Baris 2 :

-1.48 x 1 + 1.48 = 0

-1.48 x 0.59 + 0.93 = 0.057

-1.48 x 1.8 + (-1.3) = -3.964

-1.48 x 0.019 + 0.64 = 0.612

[ 1 0. 59 1.800 0.057 −3.964

2.68 3. 04 −1. 48 ][ 0.01 90.612−0.53]

Baris 3 :

-2.68 x 1 + 2.68 = 0

-2.68 x 0.59 + 3.04 = 1.46

-2.68 x 1.8 + (-1.48) = -6.304

-2.68 x 0.019 + (-0.53) = -0.58

[1 0.59 1.800 0 .057 −3.9640 1.46 −6.304 ][ 0. 019

0. 612−0.58]

Baris 2 :

0.057 : 0.057 = 1

-3.964 : 0.057 = -69.54

0.612 : 0.057 = 10.74

[1 0.59 1.800 1 −69.540 1.46 −6.304] [ 0.019

10.74−0.58]

Baris 3 :

-1.46 x 1 + 1.46 = 0

-1.46 x (-69.54) + (-6.304) = 95.2

-1.46 x 10.74 + (-0.58) = -16.2

[1 0.59 1.800 1 −69.540 0 95.2 ][ 0.019

10.74−16.2]

Baris 3 :

95.2 : 95.2 = 1

-16.2 : 95.2 = -0.17

[1 0.59 1.800 1 −69.540 0 1 ][ 0.019

10.74−0.17 ]

Baris 2 :

69.54 x 1 + (-69.54) = 0

69.54 x (-0.17) + 10.74 = -1.08

[1 0.59 1.800 1 00 0 1 ] [ 0.019

−1.08−0.17]

Baris 1 :

-1.80 x 1 + 1.80 = 0

-1.80 x (-0.17) + 0.019 = 0.32

[1 0.59 00 1 00 0 1 ][ 0. 32

−1.08−0.17]

Baris 1 :

-0.59 x 1 + 0.59 = 0

-0.59 x (-1.08) + 0.32 = 0.95

[1 0 00 1 00 0 1 ][ 0. 95

−1.08−0.17]

Jadi = 0.95X1 – 1.08X2 – 0.17X3