metode eliminasi gauss jordan
TRANSCRIPT
![Page 1: METODE ELIMINASI GAUSS JORDAN](https://reader031.vdocuments.net/reader031/viewer/2022012221/61e02aa157fb02632f61e45c/html5/thumbnails/1.jpg)
Yan Batara Putra S.Si M.Si
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Metode ini merupakan pengembangan metode eliminasi Gauss, hanya saja augmented matrik yang pada metode Eliminasi Gauss diubah menjadi matrik segitiga, pada metode Eliminasi Gauss Jordan diubah menjadi matrik diagonal.
a11 a12 a13 ... a1n b1
0 a22 a23 ... a2n b2
0 0 a33 ... a3n b3
... ... ... ... ... ...0 0 0 0 ann bn
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1 0 0 0 0 b1
0 1 0 0 0 b2
0 0 1 0 0 b3
... ... ... ... ... ...0 0 0 0 1 bn
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Matrik segitiga Matrik diagonal
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Langkah2 Metode Eliminasi Gauss Jordan 1. Buat matrik augmented 2. Buat matrik diagonal
3. Penyelesaian dari persamaan linier simultan diatas adalah nilaid1,d2,d3,…,dn dan atau:
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nd
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1...000
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0...100
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nn dxdxdxdx ,....,,, 332211
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Selesaikan persamaan linier simultan:
Dengan cara eliminasi biasa :
x1 + x2 = 3 * 2 2x1 + 2x2 = 6 substitusi2x1 + 4x2 = 8 * 1 2x1 + 4x2 = 8 x1 + x2
= 3-2x2 = -2 x1 + 1
= 3x2 = 1 x1 = 2
842
3
21
21
xx
xx
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Selesaikan persamaan linier simultan:
Dengan cara eliminasi Gauss :Augmented Matrik
Matrik Segitiga dari baris terakhir :1 1 3 B2-2B1 1 1 3 2x2 = 22 4 8 2–2(1)=0 0 2 2 x2 = 1
4–2(1)=2 substitusi, dari baris 1 :
8-2(3)=2 x1 + x2 = 3x1 + 1 = 3x1 = 2
842
3
21
21
xx
xx
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Selesaikan persamaan linier simultan:
Dengan cara eliminasi Gauss :Augmented Matrik
Matrik Segitiga 1 1 3 B2-2B1 1 1 3 B2/2 1 1 3 2/2
=12 4 8 2–2(1)=0 0 2 2 0 1 1 2/2
=14–2(1)=2 matrik diagonal8-2(3)=2 B1-B2 1 0 2 1-
1=0 0 1 1 3-
1=2 jadi x1 = 2 dan x2 = 1
842
3
21
21
xx
xx
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0563
17720
9211
B2-2B1
0563
1342
92
zyx
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Augmented matrik
1 1 2 92 4 3 13 6 5 0
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2-2(1)=04-2(1)=2-3-2(2)=-71-2(9)=-17
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B3-3B1
3-3(1)=06-3(1)=3-5-3(2)=-110-3(9)=-27
1 1 2 90 2 7 170 3 11 27
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1 1 2 90 2 7 170 0 1 3
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2B3-3B2
0563
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92
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2(0)-3(0)=02(3)-3(2)=02(-11)-3(-7)=-12(-27)-3(-17)=-3
1 1 2 90 2 7 170 0 1 3
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B3 * -1
1 1 2 90 2 7 170 3 11 27
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1 1 2 90 2 7 170 0 1 3
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B2/2
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92
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B1 – B2
1 1 2 90 2 7 170 0 1 3
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1 1 2 90 1 7 / 2 17 / 20 0 1 3
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1 1 2 90 1 7 / 2 17 / 20 0 1 3
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1 0 11/ 2 35 / 20 1 7 / 2 17 / 20 0 1 3
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1-0=11-1=02-(-7/2)=11/29-(-17/2)=35/2
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B1- 11/2 (B3)
0563
1342
92
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B2 + 7/2 (B3)
1 0 11/ 2 35 / 20 1 7 / 2 17 / 20 0 1 3
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1- 11/2 (0) =10- 11/2 (0) = 011/2 – 11/2 (1)=035/2- 11/2 (3)=1
1 0 0 10 1 7 / 2 17 / 20 0 1 3
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1 0 0 10 1 7 / 2 17 / 20 0 1 3
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0+ 7/2 (0) =01+ 7/2 (0) =1-7/2 + 7/2 (1)=0-17/2+ 7/2 (3)=4/2
1 0 0 10 1 0 20 0 1 3
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0563
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1 0 0 10 1 0 20 0 1 3
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Matrik Diagonal
Jadi :x = 1, y = 2 dan z = 3
Coba dimasukkan ke soal :1 + 2 + 2(3) = 92(1) + 4(2) - 3(3) = 13(1) + 6(2) – 5(3) = 0
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