tugas metode numerik matriks
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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
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