numeric methode

8/21/2019 numeric methode

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Systems of Linear Equations:

An Introduction

11 22 33 44 55 66

66

55

44

33

22

11

– – 11

y y

x x

(2, 3)(2, 3)

2 1 x y 2 1 x y

3 2 12 x y 3 2 12 x y


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Systems of Equations

Recall that a

system of two linear equations in twovariables

may be written in the general form

where

a

,

b

,

c

,

d

,

h

, and

k

are

real numbers

and neithera

and

b

nor

c

and

d

are both zero.

Recall that the graph of each equation in the system is a

straight line

in the plane, so that geometrically, the

solution

to the system is the

point(s) of intersection

of thetwo straight lines

L

1

and

L

2

, represented by the first and

second equations of the system.

ax by h

cx dy k


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Given the two straight lines

L

1

and

L

2

,

one and only one

ofthe following may occur:

1.

L

1

and

L

2

intersect at

exactly one point

.

y

x

L

1

L

2

Unique

solution(

x

1

,

y

1

)(

x

1

,

y

1

)

x

1

y

1


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L

1

and

L

2

,

one and only one


2.

L

1

and

L

2

are

coincident

.

y

x

L

1

,

L

2

Infinitely

many

solutions


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L

1

and

L

2

,

one and only one


3.

L

1

and

L

2

are

parallel

.

y

x

L

1

L

2

No

solution


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Example:

A System of Equations With Exactly One Solution

Consider the system

Solving

the

first equation

for

y

in terms of

x

, we obtain

Substituting

this expression for

y

into the

second equation

yields

2 1

3 2 12

x y

x y

2 1 y x

3 2(2 1) 12

3 4 2 12

7 14

2

x x

x x

x

x


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Example:


Finally, substituting this value of x into the expression for y

obtained earlier gives

Therefore, the unique solution of the system is given by

x = 2 and y = 3.

2 1

2(2) 1

3

y x


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1 2 3 4 5 6

6

5

4

3

2

1

– 1

Example:


Geometrically, the two lines represented by the two

equations that make up the system intersect at the

point (2, 3):

y

x

(2, 3)

2 1 x y

3 2 12 x y


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Example:

A System of Equations With Infinitely Many Solutions

Consider the system

Solving the first equation for y in terms of x, we obtain

Substituting this expression for y into the second equationyields

which is a true statement.

This result follows from the fact that the second equation

is equivalent to the first.

2 1

6 3 3

x y

x y

2 1 y x

6 3(2 1) 3

6 6 3 3

0 0

x x

x x


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Example:


Thus, any order pair of numbers ( x, y) satisfying theequation y = 2 x – 1 constitutes a solution to the system.

By assigning the value t to x, where t is any real number,

we find that y = 2t – 1 and so the ordered pair (t , 2t – 1)

is a solution to the system. The variable t is called a parameter.

For example:

✦ Setting t = 0, gives the point (0, – 1) as a solution of the

system.

✦ Setting t = 1, gives the point (1, 1) as another solution of

the system.


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6

5

4

3

2

1

– 11 2 3 4 5 6

Example:


Since t represents any real number, there are infinitely

many solutions of the system.

Geometrically, the two equations in the system representthe same line, and all solutions of the system are pointslying on the line:

y

x

2 1

6 3 3

x y

x y


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Example:

A System of Equations That Has No Solution

Consider the system

Solving the first equation for y in terms of x, we obtain

Substituting this expression for y into the second equationyields

which is clearly impossible.

Thus, there is no solution to the system of equations.

2 1

6 3 12

x y

x y

2 1 y x

6 3(2 1) 12

6 6 3 12

0 9

x x

x x


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1 2 3 4 5 6

Example:

A System of Equations That Has No Solution

To interpret the situation geometrically, cast both

equations in the slope-intercept form, obtaining

y = 2 x – 1 and y = 2 x – 4

which shows that the lines are parallel.

Graphically:

6

5

4

3

2

1

– 1

y

x

2 1 x y

6 3 12 x y


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Systems of Linear Equations:

Unique Solutions

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1


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The Gauss-Jordan Method

The Gauss-Jordan elimination method is a technique forsolving systems of linear equations of any size.

The operations of the Gauss-Jordan method are

1. Interchange any two equations.

2. Replace an equation by a nonzero constant multiple of

itself.

3. Replace an equation by the sum of that equation and aconstant multiple of any other equation.


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Example

Solve the following system of equations:

Solution First, we transform this system into an equivalent system

in which the coefficient of x in the first equation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

Multiply the

equation by 1/2


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Example


Solution

First, we transform this system into an equivalent systemin which the coefficient of x in the first equation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

Multiply the first

equation by 1/2


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Example


Solution

Next, we eliminate the variable x from all equations exceptthe first:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z

Replace by the sum of

– 3 X the first equation

+ the second equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z


– 3 ☓ the first equation

+ the second equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

2 2

x y z

y z

x y z

Replace by the sumof the first equation

+ the third equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Replace by the sum

of the first equation

+ the third equation


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Example


Solution

Then we transform so that the coefficient of y in thesecond equation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 4 6

3 5 13

x y z

y z

y z

Multiply the second

equation by 1/2


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Example


Solution

Then we transform so that the coefficient of y in thesecond equation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z

Multiply the second

equation by 1/2


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Example


Solution

We now eliminate y from all equations except the second:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 3 11

2 3

3 5 13

x y z

y z

y z


the first equation +

( – 2) ☓ the second equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

3 5 13

x z

y z

y z





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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

3 5 13

x z

y z

y z


the third equation +



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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

11 22

x z

y z

z


the third equation +



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Example


Solution

Now we transform so that the coefficient of z in the thirdequation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

11 22

x z

y z

z

Multiply the third

equation by 1/11


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Example


Solution

Now we transform so that the coefficient of z in the thirdequation is 1:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

7 17

2 3

2

x z

y z

z

Multiply the third

equation by 1/11


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Example


Solution

We now eliminate z from all equations except the third:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z



( – 7) ☓ the third equation

7 17

2 3

2

x z

y z

z


31/60

Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

2 3

2

x

y z

z


the first equation +( – 7) ☓ the third equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

2 3

2

x

y z

z


the second equation +

2 ☓ the third equation


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Example


Solution


2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

1

2

x

y

z


the second equation +

2 ☓ the third equation


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Example


Solution

Thus, the solution to the system is x = 3, y = 1, and z = 2.

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

3

1

2

x

y

z


35/60

Augmented Matrices

Matrices are rectangular arrays of numbers that can aid

us by eliminating the need to write the variables at eachstep of the reduction.

For example, the system

may be represented by the augmented matrixCoefficient

Matrix

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

1 1 2 2


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Matrices and Gauss-Jordan

Every step in the Gauss-Jordan elimination method can be

expressed with matrices, rather than systems of equations,

thus simplifying the whole process:

Steps expressed as systems of equations:

Steps expressed as augmented matrices:

2 4 6 22

3 8 5 27

2 2

x y z

x y z

x y z

2 4 6 22

3 8 5 27

1 1 2 2


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1 2 3 11

3 8 5 27

1 1 2 2

2 3 11

3 8 5 27

2 2

x y z

x y z

x y z


38/60







1 2 3 11

0 2 4 6

1 1 2 2

2 3 11

2 4 6

2 2

x y z

y z

x y z


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1 2 3 11

0 2 4 6

0 3 5 13

2 3 11

2 4 6

3 5 13

x y z

y z

y z


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1 2 3 11

0 1 2 3

0 3 5 13

2 3 11

2 3

3 5 13

x y z

y z

y z


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1 0 7 17

0 1 2 3

0 3 5 13

7 17

2 3

3 5 13

x z

y z

y z


42/60







1 0 7 17

0 1 2 3

0 0 11 22

7 17

2 3

11 22

x z

y z

z


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1 0 7 17

0 1 2 3

0 0 1 2

7 17

2 3

2

x z

y z

z

i G


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1 0 0 3

0 1 2 3

0 0 1 2

3

2 3

2

x

y z

z

M i d G

J d


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1 0 0 3

0 1 0 1

0 0 1 2

3

1

2

x

y

z

Row Reduced Form

of the Matrix


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Row-Reduced Form of a Matrix

Each row consisting entirely of zeros lies below all

rows having nonzero entries.

The first nonzero entry in each nonzero row is 1

(called a leading 1).

In any two successive (nonzero) rows, the leading 1

in the lower row lies to the right of the leading 1 in

the upper row.

If a column contains a leading 1, then the otherentries in that column are zeros.


47/60

Row Operations

1. Interchange any two rows.

2. Replace any row by a nonzero constant

multiple of itself.

3. Replace any row by the sum of that rowand a constant multiple of any other row.


48/60

Terminology for the

Gauss-Jordan Elimination Method

Unit Column

A column in a coefficient matrix is in unit form

if one of the entries in the column is a 1 and the

other entries are zeros.

Pivoting

The sequence of row operations that transforms

a given column in an augmented matrix into a

unit column.


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Notation for Row Operations

Letting Ri denote the ith row of a matrix, we write

Operation 1: Ri ↔ R j to mean:

Interchange row i with row j.

Operation 2: cRi to mean:

replace row i with c times row i.

Operation 3: Ri + aR j to mean:

Replace row i with the sum of row i

and a times row j.


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Example

Pivot the matrix about the circled element

Solution

3 5 9

2 3 5

3 5 9

2 3 5

1

13 R 5

3 31

52 3

2 12 R R

5

3

1

3

1 3

0 1


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The Gauss-Jordan Elimination Method

1. Write the augmented matrix corresponding tothe linear system.

2. Interchange rows, if necessary, to obtain an

augmented matrix in which the first entry in

the first row is nonzero. Then pivot the matrixabout this entry.

3. Interchange the second row with any row below

it, if necessary, to obtain an augmented matrix

in which the second entry in the second row is

nonzero. Pivot the matrix about this entry.

4. Continue until the final matrix is in row-

reduced form.


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Example

Use the Gauss-Jordan elimination method to solve the

system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 1 3

1 2 3 8

1 2 R R


53/60

Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

2 2 1 3

1 2 3 8

2 12 R R

3 1 R R

1 2 R R


54/60

Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 2 19 27

0 2 12 4

2 12 R R

3 1 R R

2 3 R R


55/60

Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

2 3 R R

1

22 R

1 0 9 12

0 2 12 4

0 2 19 27


56/60

Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1

22 R

1 0 9 12

0 1 6 2

0 2 19 27

3 2 R R


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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 31 31

3 2 R R

1

331 R


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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 9 12

0 1 6 2

0 0 1 1

1

331

R

1 39 R R

2 36 R R


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Example


system of equations

Solution

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

2 36 R R

1 39 R R

E l


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Example


system of equations

Solution

The solution to the system is thus x = 3, y = 4, and z = 1.

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

1 0 0 3

0 1 0 4

0 0 1 1

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