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
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.