REVIEW OPERASI MATRIKS T E K N I K L I N G K U N G A N I T B

REVIEW OPERASI MATRIKS

TEKNIK LINGKUNGAN ITB

MENGHITUNG INVERS MATRIKS

 c11 c12   a11 a12   c     21 c22  a21 a22 

1 0 0 1   

c11a11  c12a21  c a   c a 12 22  11 12   c21a11  c22a21    c21a12  c22a22  

1  0    0    1 

DETERMINAN

 Hanya untuk square matrices

a det  c

b a   d c

b  ad  bc 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

No solution Det [A] = 0, but system is inconsistent Then this system of equations is not solvable

1 2  x1  4  2 4  x   5    2   

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   x   0  2     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  a12x2  a13x3  ...  a1n xn  b1

a21x1  a22x2  a23x3  ...  a2n xn  b2 . . .

. . .

an1x1  an 2 x2  an3 x3  ...  ann xn  bn

CONTOH 2 x1  3x2  2 x3  x4  2 2 x1  5 x2  3x3  x4  7

matriks input

 2 x1  x2  3x3  2 x4  1  5 x1  2 x2  x3  3x4  8

2 2   2   5

3 2 5 3 1 2

3 1

 1  2  1 7 2 1   3 8

FORWARD ELIMINATION 2 2   2   5

3 2 5 3 1 2

3 1

 1  2 1 7  2 1   3 8

1 3 1  1  1 2 2   1 2 9 0 2 0 4 1  3  1   19 6 1 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   3 0 19 2  6 1 2  1 3 1  1 1  2 2   9  1 1 0 1  2 2   0 0 3  7  19   0 0  5  9  159  4 4 

FORWARD ELIMINATION 1 3 1  1 1  2 2   9  1 1 0 1  2 2   0 0 3  7  19   0 0  5  9  159  4 4 

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 1 0 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:

10 x1  7 x2  7 6 x3  2.099 x2  3x1  3.901 5 x1  x2  5 x3  6 - Kemungkinan error karena round-off (kesalahan pembulatan)

CONTOH Dari sistem persamaan linear  7 0  x1   7   10  3 2.099 6     3 . 901 x =    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  10  0  0.001  6    0 0 15005

 x1   7   x  =  6.001    2   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: a pk

akk , ak 1,k ,................, ank

Jika nilai maksimumnya Maka tukar baris p dan k.

k  p  n,

Pada baris ke p,

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 mutlak yang dipakai untuk forward elimination (pivoting element) adalah selalu paling besar

PARTIAL PIVOTING: EXAMPLE Consider the system of equations

10 x1  7 x2  7  3x1  2.099 x2  6 x3  3.901 5 x1  x2  5 x3  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 10 7 0 x 7  0   0

 1        2.5 5  x2    2.5   0.001 6  x3  6.001

Performing the row swap

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

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 x2 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