Transcript Review Operasi Matriks Menghitung invers matriks? Determinan? Matriks Singular? Menghitung invers matriks  c11 c12   a11 a12   c     21 c22  a21

Review Operasi Matriks
Menghitung invers matriks?
Determinan?
Matriks Singular?
Menghitung invers matriks
 c11 c12   a11 a12 

c



 21 c22  a21 a22 
c11a11
c a
 11 12



c12 a21
c12 a22
c21a11
c21a12
1 0
0 1 





c22 a21 

c22 a22 
1 
0 
 
0 
 
1 
Determinan
 Hanya untuk square matrices
a b  a b
det 

 ad  bc

c d  c d
a1 a2 a3 
det b1 b2 b3 
 c1 c2 c3 
 a1b 2 c3  a1b 3 c2  a2b 3 c1  a2b1 c3  a3b1 c2  a3b 2 c1
Jika determinan = 0  matriks singular,
tidak punya invers
Cari invers nya…
2 4
2 5


1 2 
2 4


Sistem Persamaan Linear
Simultaneous Linear Equations
Metode Penyelesaian
•
•
•
•
•
Metode grafik
Eliminasi Gauss
Metode Gauss – Jourdan
Metode Gauss – Seidel
LU decomposition
Metode Grafik
1 2   x1  4
1  1  x   2

 2   
Det{A}  0  A
is nonsingular
so invertible
Unique
solution
2
-2
Sistem persamaan yang tak
terselesaikan
1 2  x1  4
 2 4  x   5 

 2   
No solution
Det [A] = 0,
but system is inconsistent
Then this
system of
equations is
not solvable
Sistem dengan solusi tak terbatas
Det{A} = 0  A is singular
infinite number of solutions
1 2  x1  4
 2 4   x   8 

 2   
Consistent so solvable
Ill-conditioned system of equations
A linear system
of equations is
said to be “illconditioned” if
the coefficient
matrix tends to
be singular
Ill-conditioned system of
equations
• A small deviation in the entries of A matrix,
causes a large deviation in the solution.
2   x1   3 
 x1  1
 1
0.48 0.99  x   1.47   x   1
 2  

 2  

 x1  3
2   x1   3 
 1
 




0.49 0.99  x  1.47
x2  0 


 2  

Gaussian Elimination
Merupakan salah satu teknik paling populer
dalam menyelesaikan sistem persamaan
linear dalam bentuk:
AX  C
Terdiri dari dua step
1. Forward Elimination of Unknowns.
2. Back Substitution
Forward Elimination
Tujuan Forward Elimination adalah untuk
membentuk matriks koefisien menjadi Upper
Triangular Matrix
5
1 
 25 5 1 25
 64 8 1   0  4.8  1.56

 

144 12 1  0
0
0.7 
Forward Elimination
Persamaan linear
n persamaan dengan n variabel yang tak diketahui
a11x1  a12 x2  a13 x3  ...  a1n xn  b1
a21x1  a22 x2  a23 x3  ...  a2n xn  b2
.
.
.
.
.
.
an1x1  an2 x2  an3 x3  ...  ann xn  bn
Contoh
2 x1  3x2  2 x3  x4  2
2 x1  5 x2  3x3  x4  7
 2 x1  x2  3x3  2 x4  1
matriks input
 5 x1  2 x2  x3  3x4  8
2
2

 2

 5
3  2  1  2

5 3 1
7
1 3 2 1 

2 1
3
8
Forward Elimination
2
2

 2

 5
3  2  1  2
5 3 1
7 
1 3 2 1 

2 1
3
8
1 3
1  1
 1
2
2


1
2
9
0 2
0 4
1
 3  1


19
1
6
3
0
2
2

R1'  R1
2
R2'  R2  2 R1'
R3'  R3  2 R1'
R4'  R4  5 R1'
R1'  R1
R2'  R2
2
R3'  R3  4 R2'
R4'  R4  19 R1'
2
1 3
1  1
 1
2
2


1
2
9
0 2
0 4
1
 3  1


19
1
6
3
0
2
2

1

0

0
0

1  1
1 
2
2

9

1
1
1
2
2 
0
3
7
 19 
0 5
 9  159 
4
4
3
Forward Elimination
1

0

0
0

1
2
1 1
2
0
3
0 5
4
3
1 
2

9
1
2 
7
 19 
 9  159 
4
1
R1'  R1
R2'  R2
R3' 
R3
3
R4'  R4  5 R3'
4
1 3
1
1
1 
2
2


9

1
1
0 1

2
2



7

19
1
0 0
3
3 

143
572 
0
0
0



12
12
R1'  R1
R2'  R2
R3'  R3
R4' 
R4
143 12
1 3
1
1
1 
2
2


9
1
0 1  1 2

2



7

19
1
0 0
3
3 

143
572 
0
0
0



12
12
1 3
1  1
1 
2
2


9

1
1
0 1

2
2



7

19
0
0
1

3
3 

572 
0
0
0
1

143
Back substitution
x1 
x4  4
3 x
2 2
x2
 x3
 1 x3
2
x3
 1 x4
2
 x4
 7 x4
3
x4
 1
9
2
 19
3
 572
143
Gauss - Jourdan
1 3 2 15
2 4 3 22


3 4 7 39
2 15 
1 3
0  2  1  8 


0  5 1  6
1 0 1 2 15
0 1 1 2 4 


0 0 7 2 14
R1'  R1
R2'  R2  2 R1'
R3'  R3  3R1'
R1'  R1  3R2'
R2'   R2 2
R3'  R3  5 R2'
R1'  R1  1 R3'
2
R2'  R2  1 R3'
2
R
R3'  3 7 2
2 15 
1 3
0  2  1  8 


0  5 1  6
1 0 1 2 15
0 1 1 2 4 


0 0 7 2 14
1 0 0 1 
0 1 0 2 


0 0 1 4
Warning..
Dua kemungkinan kesalahan
-Pembagian dengan nol mungkin terjadi pada langkah
forward elimination. Misalkan:
10x1  7 x2  7
6 x3  2.099x2  3x1  3.901
5x1  x2  5x3  6
- Kemungkinan error karena round-off (kesalahan pembulatan)
Contoh
Dari sistem persamaan linear
 7 0  x1   7 
 10
 3 2.099 6   

x
3
.
901
=

  2 

 5
 1 5  x3   6 
7 0 7
 10

 3 2.099 6 3.901


 5

1 5 6
Akhir dari Forward Elimination
7
0   x1   7 
10
 0  0.001
    6.001
6


  x2  = 
 0
0
15005  x3  15004
 
7
0
7
10

 0  0.001

6
6
.
001


 0
0
15005 15004
Kesalahan yang mungkin terjadi
Back Substitution
7
0   x1   7 
10
 0  0.001
  x    6.001
6

 2  

 0
0
15005  x3  15004
x3 
15004
 0.99993
15005
6.001  6 x3
x2 
 1.5
 0.001
7  7 x 2  0 x3
x1 
 0.3500
10
Contoh kesalahan
Bandung-kan solusi exact dengan hasil perhitungan
X  exact
X  calculated
 x1   0 




  x 2    1
 x3   1 
 x1    0.35 
  x 2     1.5 
 x3  0.99993
Improvements
Menambah jumlah angka penting
Mengurangi round-off error (kesalahan pembulatan)
Tidak menghindarkan pembagian dengan nol
Gaussian Elimination with Partial Pivoting
Menghindarkan pembagian dengan nol
Mengurangi round-off error
Pivoting
Eliminasi Gauss dengan partial pivoting mengubah tata urutan
baris untuk bisa mengaplikasikan Eliminasi Gauss secara Normal
How?
Di awal sebelum langkah ke-k pada forward elimination, temukan angka
maksimum dari:
akk , ak 1,k ,................, ank
Jika nilai maksimumnya
Maka tukar baris p dan k.
a pk
Pada baris ke p,
k  p  n,
Partial Pivoting
What does it Mean?
Gaussian Elimination with Partial Pivoting ensures that
each step of Forward Elimination is performed with the
pivoting element |akk| having the largest absolute value.
Jadi,
Kita mengecek pada setiap langkah apakah angka
paling atas (pivoting element) adalah selalu paling
besar
Partial Pivoting: Example
Consider the system of equations
10x1  7 x2  7
 3x1  2.099x2  6 x3  3.901
5x1  x2  5x3  6
In matrix form
 7 0  x1 
 7 
 10
3.901
 3 2.099 6  x 


  2 = 
 6 
 5
 1 5  x 3 
Solve using Gaussian Elimination with Partial Pivoting using five
significant digits with chopping
Partial Pivoting: Example
Forward Elimination: Step 1
Examining the values of the first column
|10|, |-3|, and |5| or 10, 3, and 5
The largest absolute value is 10, which means, to follow the
rules of Partial Pivoting, we don’t need to switch the rows
Performing Forward Elimination
 7 0  x1   7 
 10
 3 2.099 6  x   3.901

 2  

 5
 1 5  x3   6 

7
0  x1   7 
10
 0  0.001 6  x   6.001

 2  

 0
2.5
5  x3   2.5 
Partial Pivoting: Example
Forward Elimination: Step 2
Examining the values of the first column
|-0.001| and |2.5| or 0.0001 and 2.5
The largest absolute value is 2.5, so row 2 is switched with
row 3
Performing the row swap
7
0  x1   7 
10
 0  0.001 6  x   6.001

 2  

 0
2.5
5  x3   2.5 

7
0  x1   7 
10
0
  x    2.5 
2
.
5
5

 2  

 0  0.001 6  x3  6.001
Partial Pivoting: Example
Forward Elimination: Step 2
Performing the Forward Elimination results in:
0   x1   7 
10  7
 0 2. 5
  x    2.5 
5

 2  

 0
0 6.002  x3  6.002
Partial Pivoting: Example
Back Substitution
Solving the equations through back substitution
0   x1   7 
10  7
 0 2. 5
  x    2.5 
5

 2  

 0
0 6.002  x3  6.002
6.002
x3 
1
6.002
2.5  5 x 2
x2 
1
2.5
7  7 x 2  0 x3
x1 
0
10
Partial Pivoting: Example
Compare the calculated and exact solution
The fact that they are equal is coincidence, but it does
illustrate the advantage of Partial Pivoting
 x1   0 
X  calculated   x2    1
 x3   1 
X  exact
 x1   0 
  x 2    1
 x3   1 
Summary
-Forward Elimination
-Back Substitution
-Pitfalls
-Improvements
-Partial Pivoting