# Chapters 9-10

1

CHAPTER 9

9.4

The equations can be rearranged into a format for plotting

x

2

versus

x

1

:

10

8

6

4

2

0

0 2 4 6 8 10

Therefore, the solution is

x

1

= 4,

x

2

= 5.

9.5

(a)

The equations can be rearranged into a format for plotting

x

2

versus

x

1

:

120

100

80

60

40

20

0

0 200 400 600 800

If you zoom in, it appears that there is a root at about (404.6, 56.5).

56.6

56.55

56.5

56.45

56.4

404 404.5

405 405.5

(b)

The plot suggests that the system may be ill-conditioned because the slopes are so similar.

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2

(c)

The determinant can be computed as

D

= −

1 .

1 ( 17 .

4 )

10 (

2 )

=

0 .

86 which is relatively small. Note that if the system is normalized first by dividing each equation by the largest coefficient,

0 .

11

x

1

+

0 .

11494

x x

1

2

+

=

x

2

12

=

10 the determinant is even smaller

D

= −

0 .

11 ( 1 )

1 (

0 .

11494 )

=

0 .

00494

(d)

Using Eqs. (9.10) and (9.11) yields

x

1

=

17 .

4 ( 120 )

0 .

86

10 ( 174 )

=

404 .

6512

x

2

=

1 .

1 ( 174 )

(

0 .

86

2 )( 120 )

= 56 .

51163

9.7

(a)

The equations can be rearranged into a format for plotting

x

2

versus

x

1

:

14.6

14.55

14.5

14.45

14.4

9.8

9.9

10 10.1

10.2

The solution is

x

1

= 10,

x

2

= 14.5. Notice that the lines have very similar slopes.

(b)

The determinant can be computed as

D

=

0 .

5 (

2 )

(

1 ) 1 .

02

=

0 .

02

(c)

The plot and the low value of the determinant both suggest that the system is illconditioned.

(d)

Using Eqs. (9.10) and (9.11) yields

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3

x

1

=

9 .

5 (

2 )

(

1 )(

18 .

8 )

=

10

0 .

02

x

2

=

0 .

5 (

18 .

8 )

0

(

.

02

9 .

5 ) 1 .

02

=

14 .

5

(e)

Using Eqs. (9.10) and (9.11) yields

x

1

=

9 .

5 (

2 )

0 .

(

02

1 )(

18 .

8 )

= −

10

x

2

=

0 .

52 (

18 .

8

)

0 .

(

02

9 .

5 ) 1 .

02

=

4 .

3

The ill-conditioned nature of the system is illustrated by the fact that a small change in one of the coefficients results in a huge change in the results.

9.8

(a)

After elimination:

 10

0

0

x

3

=

2

5

0

.

4

5

1 .

1

7

.

351852

Back substitution:

5 .

32 .

1111

351852

= −

6

x

2

=

53 .

4

5 .

1

4

.

7 (

6 )

=

8

x

1

=

27

(

1 )(

6

10

)

9.9

(a)

After elimination:

2 ( 8 )

=

32

27

53

.

.

4

1111

0 .

5

 12

0

0

0 .

2

66667

0

Back substitution:

3 .

2

33333

4

x

3

=

2

4

=

0 .

5

6

2

6

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4

x

2

=

6

(

3 .

3333 ) 0 .

0 .

66667

5

= −

6 .

5

x

1

=

6

2 ( 0 .

5 )

12

2 (

6 .

5 )

=

1 .

5

9.10

(a)

The determinant can be computed as:

D

=

0 (

2 )

(

3 ) 5

+

7 (

12 )

= −

69

(b)

Cramer’s rule

=

3

2

2

D

= =

x

1

2

3

2

7

0

1

68

69

0 .

985507

x

2

=

0

1

5

2

3

2

D

7

0

1

=

101

69

x

3

=

0

1

5

2

D

3

2

2

3

2

=

63

69

(c)

After elimination:

=

1 .

463768

=

0 .

913043

5

0

0

0

2

3

4

0

7

.

6 4

2

2

.

2

Back substitution:

x

3

=

4 .

2

4 .

6

=

0 .

913043

x

2

=

2

7 ( 0

.

913043

3

)

x

1

=

2

0 ( 0 .

913043 )

5

9.11

(a)

After elimination:

=

1 .

463768

(

2 )( 1 .

463768 )

=

0 .

985507

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5

0

0

8

5

1

0

.

75

8

1

2

.

5

.

108696

Back substitution:

x

3

=

16 .

2174

8 .

108696

= −

2

x

2

=

43

(

1 .

5 )(

2 )

5 .

75

=

x

1

=

20

+

2 (

2 )

8

1 ( 8 )

=

8

4

16 .

20

43

2174

CHAPTER 10

10.1

Matrix multiplication is distributive. Perform algebra…

10.2

(a)

LU

decomposition is

[

L

]

= 

0

1

0

.

.

1

3

0 .

0

1

14815

0

0

1

[

U

]

= 

 10

0

0

2

5

0

.

4

5 .

1 .

1

7

351852

These two matrices can be multiplied to yield the original system. For example, using

MATLAB to perform the multiplication gives

>> L=[1 0 0;-.3 1 0;0.1 -.14815 1];

>> U=[10 2 -1;0 -5.4 1.7;0 0 5.351852];

>> L*U ans =

10.0000

2.0000

-1.0000

-3.0000

1.0000

-6.0000

1.0000

2.0000

5.0000

(b)

Forward substitution: [

L

]{

D

} = {

B

}

Solving yields

d

1

= 27,

d

Back substitution:

2

= –53.4, and

d

3

= –32.1111.

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6

 10

0

0

2

5

0

.

4

5 .

1 .

1

7

351852



x

1

x x

2

3



=

32

27

53

.

.

4

1111

x

3

=

32 .

1111

5 .

351852

= −

6

x

2

=

53 .

4

1

5 .

4

.

7 (

6 )

=

8

x

1

=

27

(

1 )(

6 )

10

2 ( 8 )

=

0 .

5

(c)

Forward substitution: [

L

]{

D

} = {

B

}

0

1

0 .

.

1

3

0 .

0

1

14815

0

0

1

Solving yields

d

1

= 12,

d

Back substitution:



d d d

2

1

2

3



=

12

18

6

= 21.6, and

d

3

= –4.

 10

0

0

2

5

0

.

4

5 .

1 .

1

7

351852



x

1

x x

2

3



=

12

21

.

4

6

x

3

=

4

5 .

351852

= − 0 .

7474

x

2

=

21 .

6

1 .

7 (

0 .

7474 )

5 .

4

x

1

=

12

(

1 )(

0 .

7474 )

10

10.3

(a)

LU

decomposition is

= −

4 .

23529

2 (

4 .

23529 )

[

L

]

=

0

0

.

1

.

25

25

0 .

0

1

33333

0

0

1

Forward substitution: [

L

]{

D

} = {

B

}

=

1 .

972318

[

U

]

=

8

0

0

 4

6

0

0

6

.

1

75

.

5

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7

0

0

.

1

.

25

25

0 .

0

1

33333

0

0

1

Solving yields

d

1

= 11,

d

Back substitution:



d d d

2

1

2

3



=

11

4

7

= 6.75, and

d

3

= 6.5.

 8

0

0

4

6

0

0

6

.

.

1

75

5



x

1

x

2

x

3



=

6

11

6

.

.

75

5

x

3

=

6 .

5

6 .

5

=

1

x

2

=

6 .

75

6

0 .

75 ( 1 )

=

1

x

1

=

11

(

1 )( 1 )

4 ( 1 )

=

1

8

(b)

The first column of the inverse can be computed by using [

L

]{

D

} = {

B

}

0

1

0 .

25

.

25

0 .

0

1

33333

0

0

1



d d d

1

2

3



=

1

0

0

This can be solved for

d

1 substitution

= 1,

d

2

= 0.25, and

d

3

=

0.16667. Then, we can implement back

8

0

0

4

6

0

0

6

.

.

1

75

5



x

1

x

2

x

3



=

0

0

.

.

1

25

16667 to yield the first column of the inverse

{

X

}

=

0

0

.

.

099359

0448718

0 .

025641

For the second column use {

B

}

T

= {0 1 0} which gives {

D

}

T

= {0 1 0.33333}. Back substitution then gives {

X

}

T

= {

0.073718 0.160256 0.051282}.

For the third column use {

B

}

T

= {0 0 1} which gives { gives {

X

}

T

= {0.028846

0.01923 0.153846}.

D

}

T

= {0 0 1}. Back substitution then

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8

Therefore, the matrix inverse is

[

A

]

1

= 

 0

0 .

.

099359

0

044872

.

025641

0

0

0 .

.

.

073718

160256

051282

0

0

.

028846

0 .

019231

.

153846

We can verify that this is correct by multiplying [

A

][

A

]

–1

to yield the identity matrix. Can use

10.4

MATLAB to do this.

After elimination:

[

A

]

= 

0

0 .

8

25

.

375 0

5

1

.

75

.

23913

1

2

.

5

8 .

108696

Therefore, the

LU

decomposition is

[

L

]

= 

0

0

1

.

25

.

375 0 .

0

1

23913

0

0

1

[

U

]

= 

0

0

8

5

1

0

.

75

8 .

1 .

2

5

108696

Forward substitution: [

L

]{

D

} = {

B

}. Solving yields

d

1

=

20,

d

Back substitution:

2

=

43, and

d

3

=

16.2174.

0

0

8

5

1

0

.

75

8 .

1

2

.

108696

x

3

=

16 .

2174

8 .

108696

= −

2

x

2

=

43

+

5

1

.

.

5

75

(

2 )

x

1

=

20

+

2

(

8

2 )

10.6

After elimination:

8

=

5

8

=

4



x

1

x x

2

3



=

10

0

0 .

.

1

3

0 .

2

5 .

4

148148 5 .

1 .

1

7

351852

Therefore, the

LU

decomposition is

16

20

43

.

2174

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[

L

]

= 

 1

0 .

0 .

1

3

0 .

0

1

148148

0

0

1

[

U

]

= 

 10

0

0

2

5

0

.

4

5 .

1 .

1

7

351852

The first column of the inverse can be computed by using [

L

]{

D

} = {

B

}

0

1

0 .

.

1

3

0

0

1

.

148148

0

0

1



d d d

1

2

3



=

1

0

0

This can be solved for

d

1 substitution

= 1,

d

2

= 0.3, and

d

3

=

0.055556. Then, we can implement back

 10

0

0

2

5

0

.

4

5 .

1 .

1

7

351852



x x x

1

2

3



=

0

0

1

.

3

.

055556 to yield the first column of the inverse

{

X

}

=

0 .

110727

0

0

.

.

058824

0103806

For the second column use {

B

}

T

= {0 1 0} which gives {

D

}

T

= {0 1 0.148148}. Back substitution then gives {

X

}

T

= {0.038062

0.176471 0.027682}.

For the third column use {

B

}

T

= {0 0 1} which gives { gives {

X

}

T

= {0.00692 0.058824 0.186851}.

D

}

T

= {0 0 1}. Back substitution then

Therefore, the matrix inverse is

[

A

]

1

= 

0 .

0

0

110727

.

.

058824

010381

0

0

.

0

038062

.

176471

.

027682

0 .

0 .

006920

058824

0 .

186851

We can verify that this is correct by multiplying [

A

][

A

]

–1

to yield the identity matrix.

9

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