CBE 250 Lecture 6Monday Sept. 14, 2015

Reminders: this is the only class this week.Homework 2 (last lecture) due 9/18 to my mailbox or email

Amended to delete 5.1 and add the class example PM soln(slide 5 of Lecture 5)

Exam 1 review – in class exercises on 9/23Exam 1 scheduled for 9/25

Elimination of Unknowns

Naive Gauss Elimination (Ch. 9.2)

Extension of method of elimination to large sets of equations by developing a systematic scheme or algorithm

1. Forward elimination of unknowns2. Back substitution

11313212111 bxaxaxaxa nn

22323222121 bxaxaxaxa nn

nnnnnnn bxaxaxaxa 332211

11313212111 bxaxaxaxa nn

)1()1( n

n

n

nnnbxa

''

3

'

232

'

2222bxaxaxa nn

''

3

''

3

''

33 3bxaxa nn

11

2111313212111 *

a

abxaxaxaxa nn

1

11

211

11

21313

11

21212

11

21121 b

a

axa

a

axa

a

axa

a

axa nn

22323222121 bxaxaxaxa nn

1

11

2121

11

212313

11

2123212

11

2122 )()()( b

a

abxa

a

aaxa

a

aaxa

a

aa nnn

''

3

'

232

'

2222bxaxaxa nn

Eq. #1

Eq. #1new

Eq. #2

Subtract Eq. #2 from Eq.#1new:

Define new coefficients in your new Eq.#2

)1(

)1(

n

nn

n

nn a

bx

11313212111 bxaxaxaxa nn

)1()1( n

n

n

nnnbxa

''

3

'

232

'

2222bxaxaxa nn

''

3

''

3

''

33 3bxaxa nn

1,,2,1)1(

1

)1()1(

nnifor

a

xabx

i

ii

n

ijj

i

ij

i

i

i

Pivot Equation

Pivot element

EXAMPLE:Use Naive Gauss elimination method to solve the following set of linear equations:

x1 + 5·x2 + 6·x3 = 1 7·x1 + 4·x2 + 2·x3 = -1 -3·x1 + 6·x2 + 7·x3 = 3

Drawbacks of Eliminations

• Division by zero• Round off errors • Ill conditioned systems• Singular systems

Techniques for Improving Solutions

1. Use of more significant figures2. Scaling3. Pivoting.

i. Partial pivoting. ii. Complete pivoting

Solve by Gauss Elimination with Partial Pivoting

Chapter 11.2:GAUSS SEIDEL ITERATIVE METHOD FOR THE SOLUTION OF LINEAR EQUATIONS

[A]{X} = {B}

Suppose we have:

33

23213133

22

32312122

11

31321211

a

xaxabx

a

xaxabx

a

xaxabx

Solve first equation for x1, second equation for x2, third for x3:

3

2

1

3

2

1

333231

232221

131211

b

b

b

x

x

x

aaa

aaa

aaa

GAUSS SEIDEL ITERATIVE METHOD FOR THE SOLUTION OF LINEAR EQUATIONS

3.0x1 + 0.1·x2 - 0.2·x3 = 7.85 0.1·x1 + 7.0·x2 – 0.3·x3 = -19.3 0.3·x1 - 0.2·x2 + 10.·x3 = 71.4

Example:

10

2.03.04.71

7

3.01.03.19

3

2.01.085.7

213

312

321

newnewnew

previousnewnew

previouspreviousnew

xxx

xxx

xxx

Guess x2 = x3 = 0

Rearrange:

Check if: 3,2,1iforx

xxinew

i

previousi

newi

33

23213133

22

32312122

11

31321211

a

xaxabx

a

xaxabx

a

xaxabx

Matrix Inverse Meaning (10.2.2)

Stimulus-Response Calculations: Elements of represent the response of a single part of the system to a unit stimulus of any other part of the system.