Matakuliah : METODE NUMERIK I
Tahun
: 2008
HAMPIRAN NUMERIK PENYELESAIAN
SISTEM PERSAMAAN LINIER
Pertemuan 5
Bentuk umum Sistem
Persamaan Linier
n
a
j 1
ij
X j  ci Untuk i = 1,2,3,…,n
a11 x1  a12 x 2  a13 x3    a1n x n  c1
a 21 x1  a 22 x 2  a 23 x3    a 2 n x n  c 2
a 31 x1  a32 x 2  a33 x3    a 3n x n  c3

a n1 x1  a n 2 x 2  a n 3 x3    a nn x n  c n
Bina Nusantara
 a11
a
 21
 .

 .
 .

a n1
a12
a 22
.
.
.
an2
. . . a1n   x1   c1 
. . . a 2 n   x 2  c 2 
. . . .  .   . 
.     
. . . .  .   . 
. . . .  .   . 
   
. . . a nn   x n  c n 
Metode Penyelesaian Persamaan Linier
Eliminasi Gauss
Eliminasi Gauss-Jordan
Matriks balikan
Dekomposisi LU
Iterasi Jacobi
Bina Nusantara
Iterasi Gauss-Seidel
Eliminasi Gauss
• Sistem Persamaan dengan n persamaan dan n variabel
a11 x1  a12 x 2  a13 x3    a1n x n  c1
a 21 x1  a 22 x 2  a 23 x3    a 2 n x n  c 2
a31 x1  a32 x 2  a33 x3    a3n x n  c3

a n1 x1  a n 2 x 2  a n 3 x3    a nn x n  c n
• Nilai variabel (vektor x,) pada sistem
tidak berubah jika dilakukan hal-hal
berikut:
 Kalikan atau bagikan suatu persamaan dengan konstanta
yang tidak sama dengan nol
 Ganti suatu persamaan dengan penjumlahan persamaan
tersebut dengan persamaan lainnya
Bina Nusantara
Operasi Baris Elementer
a11x1  a12 x2  a13 x3    a1n xn  b1
a21x1  a22 x2  a23 x3    a2n xn  b2
a31x1  a32 x2  a33 x3    a3n xn  b3

an1x1  an 2 x2  an3 x3    ann xn  bn
Bina Nusantara
b’k = bk (a11) – b1 (ak1)
k = 2,3,4,…,n
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x2  a33
 x3    a3 n xn  b3
a32

 xn  bn
an 2 x2  an 3 x3    ann
Membentuk Segitiga Bawah
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x2  a33
 x3    a3 n xn  b3
a32

 xn  bn
an 2 x2  an 3 x3    ann
b’k = bk (a’ 22) – b2 (ak2)
k = 3,4,…,n
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x3    a3 n xn  b3
a33

Bina Nusantara
...
ann
xn  bn...
Forward Elimination
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x3    a3 n xn  b3
a33

...
ann
xn  bn...
xn 
Bina Nusantara
bn...
...
ann
Substitusi Kembali
xn 
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x3    a3 n xn  b3
a33

...
bn
...
an
...
ann
xn  bn...
an...1,n1xn1 + an...1,n xn  bn...1
xn1 
bn...1 - an...1,n xn
an...
1,n 1
Sehingga diperoleh
b  (a12 x2  a13 x3    a1n xn )
x1  1

a11
Bina Nusantara
n
b1  (  a1k xk )
k 2
a11
Gauss Elimination
Gauss Elimination
Input
A, c , n
Forward
Elimination
Back
Substitution
Output Solution
Vector, x
stop
Bina Nusantara
Gauss Elimination
Back Substitution
xn = bn /ann
i = n-1, n-2, ..., 1
Compute last unknown
Loop backwards over all
rows except last
sum = 0
j = i+1, ..., n
Loop over all columns
to the right of the
current row
sum = sum + ai,j *x j
xi = (bi - sum)/ai,i
Bina Nusantara
Compute (i)th unknown
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x3    a3 n xn  b3
a33

...
ann
xn  bn...
Gauss Elimination
Forward Elimination
k = 1, 2, ..., n-1
i = k+1, ..., n
Bina Nusantara
Loop over all rows
in matrix, except last
Loop over all rows
below the diagonal
position (k,k)
mik = aik/akk
Compute multiplier for
row (i) and column (k)
j = k+1, ..., n
Loop over all columns
to the right of the diagonal
position (k,k)
aij = aij - mik*akj
Update coefficient
for row (i) and column (j)
bi = bi - mik*bk
Update right hand side
for row (i) and column (j)
a11x1  a12 x2  a13 x3    a1n xn  b1
 x2  a23
 x3    a2 n xn  b2
a22
 x3    a3 n xn  b3
a33

...
ann
xn  bn...
Contoh
Forward Elimination
2 x1  x2  4 x3  16
B2+(-3/2)b1
3 x1  2 x2  x3  10
x1  3 x2  3 x3  16
2 x1  x2  4 x3  16
1
x2  5 x3  -14
2
x1  3 x2  3x3  16
B3+(-1/2)b1
2 x1  x2  4 x3  16
1
x2  5 x3  -14
2
5
x2  x3  8
2
Bina Nusantara
B3+(-5)b1
2 x1  x2  4 x3  16
x2  10 x3  -28
26 x3  78
Contoh
Substitusi Balik
x3 
78
3
26
x2  - 28  10 x3
 - 28  10(3)
2
16  ( x2  4 x3 )
2
16  2  4(3)

2
1
x1 
Bina Nusantara
GAUSS-JORDAN ELIMINATON
a11x1  a12 x2  a13 x3    a1n xn  b1
a21x1  a22 x2  a23 x3    a2n xn  b2
a31x1  a32 x2  a33 x3    a3n xn  b3

an1x1  an 2 x2  an3 x3    ann xn  bn
x1  0  0    0  b1...
0  x 2  0    0  b2...
0  0  x3    0  b3...

Bina Nusantara
0  0  0  ...  0  x nn  bn...
Contoh:
Selesaikan SPL berikut dengan eliminasi Gass-Jordan
3 x1  0.1x 2  0.2 x3  7.85
0.1x1  7 x 2  0.3 x3  19.3
0.3 x1  0.2 x 2  10 x3  71.4
Jawaban:
Matriks perluasan(augmented matrix) dari koefisien SPL:
 3  0.1  0.2 7.85 
0.1 7  0.3  19.3 b


0.3  0.2 10 71.4 
Bina Nusantara
’
1
= b1/3
 1
0.1

0.3
 0.033333
 0.066667
7
 0.3
 0.2
10
2.61667 

 19.3 
71.4 
b’2 = b2 – 0.1 b1
b’3 = b3 – 0.3 b1
1
0


0
 0.033333
 0.066667
7.033333
 0.29333
 0.190000
10.0200
b”2 = b’2/7.033333
Bina Nusantara
2.61667 
 19.5617

70.6150 

1 0  0.068063 2.52356 
0 1  0.041664  2.79320


0 0
1
7.0000 
b”’1 = b”1 + 0.068063 b”’3
b”’2 = b’’2 + 0.190000 b”’3
1
0


0
Bina Nusantara
0
0
1
0
0
1
3.0000 
 2.5000 

7.0000 

 x1   3.0000 
 x    2.5000
 2 

 x3   7.0000 
1 0  0.068063 2.52356 
0 1  0.041664  2.79320


0 0
1
7.0000 
b”’1 = b”1 + 0.068063 b”’3
b”’2 = b’’2 + 0.190000 b”’3
1
0


0
Bina Nusantara
0
0
1
0
0
1
3.0000 
 2.5000 

7.0000 

 x1   3.0000 
 x    2.5000
 2 

 x3   7.0000 
Iterative Methods
•
Consider the linear system
Ax  b
 a11 a12  a1n  x1 
 b1 

 
 
a2n  x2 
 a21 a22
 b2 

 
  
  


 
 
 an1 an 2  ann  xn 
 bn 
a11x1  a12 x2    a1n xn  b1
a21x1  a22 x2    a2n xn  b2

Bina Nusantara
an1x1  an 2 x2    ann xn  bn
Solution Methods
•
•
•
•
•
Bina Nusantara
Gauss Elimination – Subject to roundoff errors and ill conditioning
Iterative methods -- Alternative to elimination method
Take initial guess of solution and then iterate to obtain improved
estimates of the solution
Jacobi and Gauss-Seidel methods
Work well for large sets of equations
Iterative Methods
•
Rearrange the equations so that an unknown is on the left-hand side of
each equation:
x1 
b1  (a12 x2    a1n xn )
a11
x2 
b2  (a21x1    a2n xn )
a22

xn 
bn  (an1x1  an 2 x2    ann 1xn 1 )
ann
•
Bina Nusantara
Initial guess
x10 , x20 ,, xn0
Jacobi Method
•
Initial guess
•
Next approximation of the solution
x10 , x20 ,, xn0
0
0
b

(
a
x



a
x
)
1
12
1
n
n
1
2
x1 
a11
0
0
b

(
a
x



a
x
)
2
21
2
n
n
1
1
x2 
a22

0
0
0
b

(
a
x

a
x



a
x
)
n
n
1
n
2
nn

1
1
1
2
n

1
xn 
ann
Bina Nusantara
Jacobi Method
•
After k iterations of this process
k
k
b

(
a
x



a
x
1
12 2
1n n )
k 1
x1 
a11
k
k
b

(
a
x



a
x
2
21 1
2n n )
k 1
x2 
a22

k
k
k
b

(
a
x

a
x



a
x
)
n
n
1
n
2
nn

1
k 1
1
2
n

1
xn 
ann
Bina Nusantara
Example – Jacobi Method
•
•
Rearrange
System of 3 equations in 3 unknowns
2 x1  x2  1
 x1  3 x2  x3  8
 x2  2 x3   5
x1k 1 
1  (  x2k )
2

1  x2k
2
k
k
k
k
8

(

x

x
)
8

x

x
1
3 
1
3
x2k 1 
3
3
x3k 1 
Bina Nusantara
 5  (  x2k )
2

 5  x2k
2
Example – Jacobi Method
•
x11 
x12 
x13 
Initial guess
1  x20
2

1 0
 0. 5
2
8  x10  x30
3
 5  x20
2
x10  0, x20  0, x30  0
1 1
8

x
8  0 0
1  x3  8  0.5  (2.5)  2

 2.667 x 2 
2
3
3
3
50

 2.5
2
• After 20 iterations
Bina Nusantara
1
1

x
1  2.6667
2
2
x1 

 1.833335
2
2
x32 
 5  x12
2
 5  1.833335

 1.1667
2
x1  2, x2  3, x3  1
Jacobi Methode
•
We stop when
 r Ai
 r A1
 r A2
~
xik 1  ~
xik

 tolerance,
k 1
~
xi
for all i
~
x1k 1  ~
x1k

 tolerance
k 1
~
x1
~
x 2k 1  ~
x 2k

 tolerance
k 1
~
x
2

 r An
Bina Nusantara
~
x nk 1  ~
x nk

 tolerance
k 1
~
xn