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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.

**PROPRIETARY MATERIAL**

. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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

**PROPRIETARY MATERIAL**

. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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

**PROPRIETARY MATERIAL**

. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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

**PROPRIETARY MATERIAL**

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.

**PROPRIETARY MATERIAL**

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

**PROPRIETARY MATERIAL**

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

**PROPRIETARY MATERIAL**

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

**PROPRIETARY MATERIAL**

[

*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

**PROPRIETARY MATERIAL**