direct methods for solving systems of linear equations

Presented by:

Acosta Creus Mileidy Lorena 2073699

Teacher:

Ph.D. Eduardo Carrillo Zambrano

Numerical Methods in Petroleum Engineering

INDUSTRIAL UNIVERSITY OF SANTANDER

2010

1. CONVENTIONAL METHODS OF SOLUTION

1.1. Mathematical background

1.2. Solving Small Numbers Of Equations

1.2.1. Graphical method

1.2.2. Cramer’s rule

1.2.3. Elimination of Unknowns

2. TECHIQUES FOR IMPROVING SOLUTIONS

2.1. Use of more significant figures

2. 2.Pivoting

2.3. Scaling

3. COMPLEMENTARY TECHNIQUES

3.1. Gauss-Jordan

3.2. LU Decomposition

4. BIBLIOGRAPHY

SCHOOL OF PETROLEUM ENGINEERING

mnmjmm

inijii

nj

nj

aaaa

aaaa

aaaa

aaaa

A

......

::::

......

::::

......

......

21

21

222221

111211

An matrix A of mxn is a

rectangular array of mn numbers

arranged in m rows and n

columns. The component or

element ij of A, denoted by aij is

the number that appears in the

row i and column j of A.

SCHOOL OF PETROLEUM ENGINEERING

TYPE DESCRIPTION EXAMPLE

SQUARE

Matrix A (mxn) where the

number of rows equals the

number of columns, ie m = n.

TRANSPOSE

To A = (aij) a matrix of m x n.

The transpose is obtained by

exchanging the rows by the

columns of A.

SYMMETRIC

It’s a square matrix that is

equal to its transpose

1 2 32 3

4 0 5 B1 5

3 1 2

A

1 2 4 61 2 4 6

2 7 3 5 2 7 3 5

4 3 8 0 4 3 8 06 5 0 4

6 5 0 4

tB B

SCHOOL OF PETROLEUM ENGINEERING

TYPE DESCRIPTION EXAMPLE

UPPER AND

LOWER

TRIANGULAR

Triangular matrix is a special

kind of square matrix whose

elements above or below its

main diagonal are zero

Uppper Lower

AUGMENTED

It’s obtained by combining two

matrices.

BANDED

It has all the elements equal to

zero, with the exception of a

band centered on the main

diagonal

11 12

21 22 23

32 33 34

43 44

a a

a a a

A

a a a

a a

SCHOOL OF PETROLEUM ENGINEERING

TYPE DESCRIPTION EXAMPLE

DIAGONAL

It’s a square matrix where all the

elements outside the main

diagonal are zero

IDENTITY

It’s a diagonal matrix where all

elements on the main diagonal

are equal to 1

11

22

33

44

0 0 0

0 0 0

0 0 0

0 0 0

a

a

A

a

a

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

A

SCHOOL OF PETROLEUM ENGINEERING

The following are several methods

that are appropriate for solving

small (n ≤3) sets of simultaneous

equations and that don’t require a

computer.

These are:

• Graphical method

• Determinants and Cramer’s rule

• Elimination of Unknowns

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

A graphical solution is obtainable for

two equations by plotting them on

Cartesian coordinates with one axis

corresponding to x1 and the other to

x2, because each equation is a straight

line.

a11x

1 + a

12x

2 = b

1

a21

x1 + a

22x

2 = b

2

Despejando x2

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

Beyond three equations, graphical methods break down and,

consequently, have little practical value for solving

simultaneous equations. There are three cases that can pose

problems when solving sets of linear equations:

x1

x2

x1

x2

x1

x2

a) No solution b) Infinite solutions c) ill-conditioned system

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

EXAMPLE:

Let x1 be the abscissa. Solve both

equations for x2:

3x1 + 2x

2 = 18

- x1 + 2x

2 = 2

𝑥2 = −3

2𝑥1 + 9

𝑥2 =1

2𝑥1 + 1

Solve:

X1=4 ; X

2=3

With the graphical method

to solve:

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

Each unknown in a system of linear

algebraic equations may be expressed as a

fraction of two determinants with

denominator D and with the numerator

obtained from D by replacing the column

of coefficients of the unknown in question

by the constants b1,

b2,…,

b

n.

For example,

FUENTE:

http://auladeblanca.blogspot.com

Gabriel Cramer

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

With Cramer’s rule to solve:

0.3x1 + 0.52x

2 + x

3= -0.01

0.5x1 + x

2 + 1.9x

3= 0.67

0.1x1 + 0.3x

2 + 0.5x

3= -0.44

Solution

D=

0.3 0.52 10.5 1 1.90.1 0.3 0.5

EX

AM

PLE

To write the determinant D: 1

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

A1=

1 1.90.3 0.5

=1(0.5)-1.9(0.3)=-0.07

A1=

0.5 1.90.1 0.5

=0.5(0.5)-1.9(0.1)=0.06

A1=

0.5 10.1 0.3

=0.5(0.3)-1(0.1)=0.05

To solve the minors: 2 Evaluating the determinant 3

D=0.3(-0.07)-0.52(0.06)+1(0.05)

D=-0.0022

These can be used to

evaluate the determinant,

so:

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

x1=

−0.01 0.52 10.67 1 1.9

−0.44 0.3 0.5−0.0022

=

0.03278

−0.0022=-14.9

x2=

0.3 −0.01 10.5 0.67 1.90.1 −0.44 0.5

−0.0022=

0.0649

−0.0022=-29.5

x3=

0.3 0.52 −0.010.5 1 0.670.1 0.3 −0.445

−0.0022=

−0.04356

−0.0022=19.8

Employing Cramer’s rule: 4

«For more than three

equations, Cramer´s rule

becomes impractcal

because, as the number of

equatioms increases, the

determinants are time

consuming to evaluate by

hand»

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

The basic strategy is to multiply the equations by constants so

that one of the unknowns will be eliminated when the two

equations are combined.

The result is a single equation that can be solved for the

remaining unknown.

a11x

1 + a

12x

2 = b

1 a

21a

11x

1 + a

21a

12x

2 = b

1a

21

a21

x1 + a

22x

2 = b

2 a

21a

11x

1 + a

22a

11x

2 = b

1a

11

3

2

5

SCHOOL OF PETROLEUM ENGINEERING

EXAMPLE:

3x1 + 2x

2 = 18

- x1 + 2x

2 = 2

𝑥1 =2 18 −2(2)

3 2 −2(−1)=4 𝑥2 =

3 2 −18(−1)

3 2 −2(−1)=3

With elimination of unknownse:

Using the last equations:

These values are equals to graphical methods.

ESCUELA DE INGENIERÍA DE PETRÓLEOS

2.1. Use of more

significant figures

• To use more significant

figures in the

computation.

2.2. Pivoting

• Problems occur

when a pivot

element is nearly or

is equal to zero.

• Partial pivoting

(Rows are switched)

• Complete partial

(Columns and rows

are exchanged)

2.3. Scaling

• It has utility in

minimizing round-off

those cases where

some of the equations

in a system have much

larger coefficients than

others.

ESCUELA DE INGENIERÍA DE PETRÓLEOS

ESCUELA DE INGENIERÍA DE PETRÓLEOS

Gaussian elimination is an algorithm of linear algebra to

determine the solutions of a system of linear equations, matrices

and inverse found.

To convert linear equation system to banded matrix 1

mnmnmm

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

.....

.....

.....

2211

22222121

11213112

mmnmm

n

n

b

b

b

aaa

aaa

aaa

2

1

21

22221

11211

.......

........

........

........_ n21 xxx

ESCUELA DE INGENIERÍA DE PETRÓLEOS

Go to the leftmost nonzero column 2

mmnmm

n

n

b

b

b

aaa

aaa

aaa

2

1

21

22221

11211

.......

........

........

........_ n21 xxx

If the first line has a zero in this column, swap it with another

that does not have

3

0,,0 1211 aa

mmnmm

n

n

b

b

b

aaa

aaa

aaa

2

1

12

22122

11112

.......

........

........

ESCUELA DE INGENIERÍA DE PETRÓLEOS

Get zeros below the pivot, adding appropriate multiples of row

than the row below it

4

12 1 1

11 12 1 1

11 11 11

21 22 2 2

21 22 2 2

11

1 2

1 2

................

........ 1........( )

.......

.......

n

n

n

n

m m mn m

m m mn m

a a b

a a a b

a a a

a a a b Fa a a b

a

a a a b

a a a b

ESCUELA DE INGENIERÍA DE PETRÓLEOS

Repeat the operations with of the other rows to obtain

the higher triangular matrix

3

1.......0........0

...1....0....0

........1...0

........1

3

213

11312

n

n

n

a

aa

aaa

Starting with the last line is not zero, move up to get

the identity matrix.

3

1.......0........0

0...1....0....0

........1...0

.........1

)(*)1(

1.......0........0

...1....0....0

........1...0

........1

223

11312

33

213

11312

n

n

nn

n

n

aa

aaa

aFnnFa

aa

aaa

ESCUELA DE INGENIERÍA DE PETRÓLEOS

Solution of the system of equations 3

bn

b

b

b

xn

x

x

x

3

2

1

.......0........0

0...3....0....0

0....0....2...0

0...0....0...1 1 1

2 2

n n

x b

x b

x b

ESCUELA DE INGENIERÍA DE PETRÓLEOS

EX

AM

PLE

322

1123

82

zyx

zyx

zyx

To find the unknowns of the linear equation system

using the method Gauss-Jordan

3212

11213

8112

1

2

F

3212

11213

42/12/11

ESCUELA DE INGENIERÍA DE PETRÓLEOS

EX

AM

PLE

3212

11213

42/12/11

312

3 21

FF

FF

5120

12/12/10

42/12/11

2*2F

5120

2110

42/12/11

ESCUELA DE INGENIERÍA DE PETRÓLEOS

EX

AM

PLE

3212

11213

42/12/11

312

3 21

FF

FF

5120

12/12/10

42/12/11

2*2F

5120

2110

42/12/11

ESCUELA DE INGENIERÍA DE PETRÓLEOS

EX

AM

PLE

2 1

2 3

(1/ 2)

2

F F

F F

1100

2110

3101

5120

2110

42/12/11

1*3

13

23

F

FF

FF

1100

3010

2001

ESCUELA DE INGENIERÍA DE PETRÓLEOS

The primary appeal of LU decomposition is that the time-

consuming elimination step can be formulated so that it involves

only operations on the matrix of coefficients (A), Thus it is well

suited for those situations where many right-hand-side vectors

(B) must be evaluated for a single value of (A).

The linear equation system can be rearranged to give 1

[A][x]=[B]

ESCUELA DE INGENIERÍA DE PETRÓLEOS

To express the linear equation system as an upper

triangular system

3

[L][D]=[B]

Assume that there is a lower diagonal matrix with 1’s

on the diagonal

2

[U][x]=[D]

𝑈 =

𝑢11 𝑢12 𝑢13

0 𝑢22 𝑢23

0 0 𝑢∗

𝑥1

𝑥2

𝑥3

=

𝑑1

𝑑2

𝑑3

L=1 0 0

𝑙12 1 0𝑙13 𝑙23 1

ESCUELA DE INGENIERÍA DE PETRÓLEOS

a) Decomposition

[A] [X]= [B]

[U] [L]

[L] [D]= [B]

[U] [X]= [D]

[X]

b) Forwards

c) Backward

Substitutio

n

Sum

mary o

f steps

[ D ]

ESCUELA DE INGENIERÍA DE PETRÓLEOS

EX

AM

PLE

Consider solving the system of equations by LU decomposition

1 2 3

1 2 3

1 2 3

2 3 4 6

4 5 10 16

4 8 2 2

x x x

x x x

x x x

Solution

2 3 4

4 5 10

4 8 2

A

1 0 0 2 3 4

2 1 0 y U= 0 1 2

2 2 1 0 0 6

L

ESCUELA DE INGENIERÍA DE PETRÓLEOS

1

2

3

2 3 4 6

0 1 2 4

0 0 6 2

x

x

x

3

3

2

3 2

1

0.333

4 23.333

1

6 4 37.333

2

x

xx

x xx

3

3

2

3 2

1

0.333

4 23.333

1

6 4 37.333

2

x

xx

x xx

7.333

3.333

0.333

x

Ux=y

So,

Finally, the solution to the linear system is given:

ESCUELA DE INGENIERÍA DE PETRÓLEOS