7/31/2019 NM_02 Linear Eq.

1/36

CHAPTER 02

PENYELESAIAN PERSAMAAN

LINEAR

7/31/2019 NM_02 Linear Eq.

2/36

2

Solving Systems of Equations

A linear equation in n variables:

a1x1 + a2x2 + + anxn = b

For small (n 3), linear algebra provides several tools to solve such

systems of linear equations:

Graphical method

Cramers rule

Method of elimination

Nowadays, easy access to computers makes the solution ofvery large

sets of linear algebraic equations possible

7/31/2019 NM_02 Linear Eq.

3/36

3

Determinants and Cramers Rule

[A] : coefficient matrix[ ]{ } { }bxA =

=

3

2

1

3

2

1

333231

232221

131211

b

b

b

x

x

x

aaa

aaa

aaa

[ ]

=

333231

232221

131211

aaa

aaa

aaa

A

D

aab

aab

aab

x33323

23222

13121

1 =D

aba

aba

aba

x33331

23221

13111

2=

D

baa

baa

baa

x33231

22221

11211

3 =

D : Determinant of A matrix

7/31/2019 NM_02 Linear Eq.

4/36

4

333231

232221

131211

aaa

aaa

aaa

D =

[ ]

=

333231

232221

131211

aaa

aaa

aaa

A

[ ]{ } { }BxA =

3231

2221

13

3331

2321

12

3332

2322

11AoftDeterminanaa

aaa

aa

aaa

aa

aaaD +==

Computing the

Determinant

23323322

3332

2322

11 aaaaaa

aaD ==

22313221

3231

222113

23313321

3331

2321

12

aaaaaa

aa

D

aaaaaa

aaD

==

==

7/31/2019 NM_02 Linear Eq.

5/36

5

Gauss Elimination

SolveAx = b

Consists of two phases:

Forward elimination

Back substitution

Forward Elimination

reducesAx = b to an uppertriangular system Tx = b

Back substitution can thensolve Tx = b forx

''

3

''

33

'

2

'

23

'

22

1131211

3333231

2232221

1131211

00

0

ba

baa

baaa

baaa

baaa

baaa

Forward

Elimination

Back

Substitution

11

21231311

'

22

3

'

23

'

22''

33

''

33

a

xaxabx

a

xabx

a

bx

=

==

7/31/2019 NM_02 Linear Eq.

6/36

6

Gaussian Elimination

Forward Elimination

x1 - x2 + x3 = 6

3x1 + 4x2 + 2x3 = 92x1 + x2 + x3 = 7

x1 - x2 + x3 = 6

0 +7x2 - x3 = -90 + 3x2 - x3 = -5

x1 - x2 + x3 = 6

0 7x2 - x3 = -9

0 0 -(4/7)x3=-(8/7)

-(3/1)

Solve using BACK SUBSTITUTION: x3 = 2 x2=-1 x1 =3

-(2/1) -(3/7)

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0 0

7/31/2019 NM_02 Linear Eq.

7/36

7

Back Substitution

1x0 +1x1 1x2 +4x3 8=

2x1 3x2 +1x3 5=

2x2 3x3 0=

2x3 4=x3 = 2

7/31/2019 NM_02 Linear Eq.

8/36

8

1x0 +1x1 1x2 0=

2x1 3x2 3=

2x2 6=

Back Substitution

x2 = 3

7/31/2019 NM_02 Linear Eq.

9/36

9

1x0 +1x1 3=

2x1 12=

Back Substitution

x1 = 6

7/31/2019 NM_02 Linear Eq.

10/36

1x0 9=

Back Substitution

x0 = 9

7/31/2019 NM_02 Linear Eq.

11/36

11

i n down to 1

/* calculatexi */

x [ i ] b [ i ]/a [ i, i ]

/* substitute in the equations above */

for j 1 to i-1 dob [j ] b [j ] x [ i ] a [j, i ]

endfor

Back Substitution(* Pseudocode *)

Time Complexity? O(n2)

7/31/2019 NM_02 Linear Eq.

12/36

12

Gaussian EliminationForward Elimination

0=

+=

ii

ji

iijijia

aaaa

7/31/2019 NM_02 Linear Eq.

13/36

13

Forward Elimination

4x0 +6x1 +2x2 2x3 = 8

2x0 +5x2 2x3 = 4

4x0 3x1 5x2 +4x3 = 1

8x0 +18x1 2x2 +3x3 = 40

-(2/4)

M

U

L

T

IP

L

I

E

R

S

-(-4/4)

-(8/4)

7/31/2019 NM_02 Linear Eq.

14/36

14

4x0 +6x1 +2x2 2x3 = 8

+4x2 1x3 = 0

+3x1 3x2 +2x3 = 9

+6x1 6x2 +7x3 = 24

3x1

-(3/-3)

M

UL

T

I

P

L

I

E

R

S

Forward Elimination

-(6/-3)

7/31/2019 NM_02 Linear Eq.

15/36

15

4x0 +6x1 +2x2 2x3 = 8

+4x2 1x3 = 0

1x2 +1x3 = 9

2x2 +5x3 = 24

3x1

??

M

U

LT

I

P

L

I

E

R

Forward Elimination

7/31/2019 NM_02 Linear Eq.

16/36

16

4x0 +6x1 +2x2 2x3 = 8

+4x2 1x3 = 0

1x2 +1x3 = 9

3x3 = 6

3x1

Forward Elimination

7/31/2019 NM_02 Linear Eq.

17/36

17

Gaussian EliminationOperation count in Forward Elimination

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0 0

2n

2n

2n

2n

2n

2n

2n

2n

TOTAL

1st column: 2n(n-1) 2n2

)(

6

)12)(1(2

2)1(*2)2(*2...)1(22

:NELIMINATIOFORWARDforOperationsof#TOTAL

3

1

22222

nO

nnn

inn

n

i

=

++=

=++++ =

2(n-1)2 2(n-2)2 .

b11

b22

b33

b44

b55

b66

b77

b66

b66

7/31/2019 NM_02 Linear Eq.

18/36

18

Pitfalls of Elimination Methods

Division by zeroIt is possible that during both elimination and back-substitution phases a

division by zero can occur.

For example:

2x2 + 3x3 = 8 0 2 3

4x1 + 6x2 + 7x3 = -3 A = 4 6 7

2x1 + x2 + 6x3 = 5 2 1 6

Solution: pivoting (to be discussed later)

7/31/2019 NM_02 Linear Eq.

19/36

19

Pitfalls (cont.)

Round-off errors

Because computers carry only a limited number of significant figures, round-off

errors will occur and they willpropagate from one iteration to the next.

This problem is especially important when large numbers of equations (100 ormore) are to be solved.

Always use double-precision numbers/arithmetic. It is slow but needed for

correctness!

It is also a good idea to substitute your results back into the original equations and

check whether a substantial error has occurred.

7/31/2019 NM_02 Linear Eq.

20/36

ill-conditioned systems - small changes in coefficients result in large changes

in the solution. Alternatively, a wide range of answers can approximately satisfythe equations.

(Well-conditioned systems small changes in coefficients result in small changes

in the solution)

Problem: Since round off errors can induce small changes in the coefficients, thesechanges can lead to large solution errors in ill-conditionedsystems.

Example:

x1 + 2x2 = 10

1.1x1 + 2x2 = 10.4

x1 + 2x2 = 10

1.05x1 + 2x2 = 10.4

34

2.0

)4.10(2)10(2

)1.1(2)2(1

24.10

210

2

222

121

1 ==

=

== xD

ab

ab

x

181.0

)4.10(2)10(2

)05.1(2)2(1

24.10

210

2

222

121

1 ==

=

== x

D

ab

ab

x

Pitfalls (cont.)

7/31/2019 NM_02 Linear Eq.

21/36

Surprisingly, substitution of the erroneous values,x1=8 andx2=1, into the originalequation will not reveal their incorrect nature clearly:

x1 + 2x2 = 10 8+2(1) = 10 (the same!)

1.1x1 + 2x2 = 10.4 1.1(8)+2(1)=10.8 (close!)

IMPORTANT OBSERVATION:

An ill-conditioned system is one with adeterminant close to zero

If determinant D=0 then there are infinitely many solutionssingular system

Scaling (multiplying the coefficients with the same value) does not change the equationsbut changes the value of the determinant in a significant way.

However, it does not change the ill-conditionedstate of the equations!

DANGER! It may hide the fact that the system is ill-conditioned!!

How can we find out whether a system is ill-conditionedor not?

Not easy! Luckily, most engineering systems yield well-conditioned results!

One way to find out: change the coefficients slightly and recompute & compare

ill-conditioned systems (cont.)

7/31/2019 NM_02 Linear Eq.

22/36

22

COMPUTING THE DETERMINANT OF A MATRIX

USING GAUSSIAN ELIMINATION

The determinant of a matrix can be found using Gaussian elimination.

Here are the rules that apply:

Interchanging two rows changes the sign of the determinant.

Multiplying a row by a scalar multiplies the determinant by that scalar.

Replacing any row by the sum of that row and any other row does NOT change the

determinant.

The determinant of a triangular matrix (upper or lower triangular) is the product of the

diagonal elements. i.e. D=t11t22t33

33

2322

131211

00

0

t

tt

ttt

D =

7/31/2019 NM_02 Linear Eq.

23/36

Techniques for Improving Solutions

Use of more significant figures double precision arithmetic

Pivoting

If a pivot element is zero, normalization step leads to division by zero. Thesame problem may arise, when the pivot element is close to zero. Problem canbe avoided:

Partial pivoting

Switching the rows below so that the largest element is the pivot element.

Go over the solution in: CHAP9e-Problem-11.doc

Complete pivoting

Searching for the largest element in all rows and columns then switching.

This is rarely used because switching columns changes the order of xsand adds significant complexity and overhead costly

Scaling

used to reduce the round-off errors and improve accuracy

7/31/2019 NM_02 Linear Eq.

24/36

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

1 1 2 1

2 1 2 1

1 2 3

2 1 2 1 3 2 3 2

2 3 3 3

3 2 3 2

2 2

ln / ln /

0ln / ln / ln / ln /

2 2

ln / ln /

s s

i i S

s s i i

i i

O O a

k kh D T T h D T

D D D D

k k k k T T T

D D D D D D D D

k kT h D T h D T

D D D D

+ =

+ + =

+ =

7/31/2019 NM_02 Linear Eq.

25/36

25

Gauss-Jordan Elimination

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0 0

0

a11

x22

0

x33

0

0

x44

0

0

0

x55

0

0

0

0

x66

x77

x88

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

b11

b22

b33

b44

b55

b66

b77

b66

b660 0 0 x99

7/31/2019 NM_02 Linear Eq.

26/36

26

Gauss-Jordan Elimination: Example

=

10

1

8

|

|

|

473

321

211

:MatrixAugmented

10

1

8

473

321

211

3

2

1

x

x

x

1 1 2 | 8

0 1 5 | 9

0 4 2| 14

R2 R2 - (-1)R1

R3 R3 - ( 3)R1

Scaling R2:

R2 R2/(-1)

R1 R1 - (1)R2

R3 R3-(4)R2

1 1 2 | 8

0 1 5| 9

0 4 2| 14

22

9

17

|

|

|

1800

510

701

Scaling R3:

R3 R3/(18)

9/11

9

17

|

|

|

100

510

701

222.1

888.2

444.8

|

|

|

100

010

001

R1 R1 - (7)R3

R2 R2-(-5)R3

RESULT:

x1=8.45, x2=-2.89, x3=1.23

Time Complexity? O(n3)

7/31/2019 NM_02 Linear Eq.

27/36

Contoh

Selesaikan persamaan berikut:

x1 + x2 + x3 +x4 +x5 = 2

x1 + 2 x3 x4 x5 = -1

2x1 x2 x3 +x4 x5 = 7

X2 x3 + 2x5 = -9

x1 + 2 x2 + x3 x4 =-2

7/31/2019 NM_02 Linear Eq.

28/36

7/31/2019 NM_02 Linear Eq.

29/36

Tugas 04 Eliminasi Gauss-Jordan

Selesaikan persamaan dalam contoh sebelumnya

memakai metode Gauss-Jordan

7/31/2019 NM_02 Linear Eq.

30/36

contoh kasus

Reaksi fase cair reaktan A menjadi B dijalankan

dalam sebuah reaktor batch

Data yang diperoleh:

T, K k, menit-1

300 0,017

340 0,17

360 0,425

380 0,955

400 2,1

440 7,81

460 13,2

Diminta mencari konstanta reaksi sebagai

fungsi suhu (T), dengan asumsi konstanta

kecepatan reaksi mengikuti persamaan :

Carilah nilai kesalahan relatif rerata

antara persamaan empiris dengan data

laboratorium!

7/31/2019 NM_02 Linear Eq.

31/36

Linearisasi

Persamaan terlinierisasi di atas dapat dianalogikan menjadi:

y= a + bx1+ cx2

dengan

Diperoleh 3 persamaan dengan 3 variabel yang tidak diketahui

Cara penentuan nilai a, b, dan ca =ln Ar Ar= e

a

b =n n = b

c = -Er Er =-c

contoh kasus

7/31/2019 NM_02 Linear Eq.

32/36

contoh kasus

7/31/2019 NM_02 Linear Eq.

33/36

Disusun menjadi matriks:

dengan elemen-elemennya adalah:

contoh kasus

7/31/2019 NM_02 Linear Eq.

34/36

Dengan melakukan eliminasi Gauss-Jordan diperoleh:

contoh kasus

a =ln Ar Ar= ea

b =n n = b

c = -Er Er =-c

7/31/2019 NM_02 Linear Eq.

35/36

contoh kasus

Tahapan pengerjaan program:

1. Input data dalam matriks

2. Pengisian dan penyusunan matriks awal

3. Eliminasi Gauss-Jordan

4. Menghitung nilai konstanta Ar, n dan Er

5. Menghitung kesalahan relatif

function latihan %eliminasi Gauss-jordan

7/31/2019 NM_02 Linear Eq.

36/36

clear;clc;

T=[300;340;360;380;400;440;460];

kdata=[0.017;0.17;0.425;0.955;2.1;7.81;13.2]

;

Ndata=7;

%Persamaan kecepatan reaksi k = Ar x T^n x

exp(-Er/T)

%dilinearisasikan menjadi bentuk y = a + b.x1

+ c.x2

%maka: ln k = ln Ar + n.ln T - Er/T

%dimana y = ln k x1 = ln T x2 = 1/T% a = ln Ar b = n c = -Er

y=log(kdata);

x1=log(T);

x2=1./T;

%pengisian matriks

A=[Ndata sum(x1) sum(x2) sum(y)

sum(x1) sum(x1.^2) sum(x1.*x2) sum(x1.*y)

sum(x2) sum(x1.*x2) sum(x2.^2) sum(x2.*y)]

j

GJ=rref(A)

%konstanta

a=GJ(1,4);b=GJ(2,4);

c=GJ(3,4);

Er=-c

Ar=exp(a)

n=b

%kesalahan relatif

kcal=Ar*(T.^n).*exp(-Er./T)

%for i=1:length(kdata)

% kcal(i,1)=Ar*(T(i)^n)*exp(-Er/T(i));

%enderror_total=sum(abs((kdata-kcal)./kdata)*100)

error_average=error_total/Ndata