Numerical Methods in Engineering with Python 3 (3rd Ed., 2013) – Solutions Manual – Kiusalaas

CHAPTER 2 - 23 COVERED




SOLUTIONS MANUAL

,PROBLEM SET 2.1

Problem 1

(a)

1 2 3
3 4 2 4 2 3
jAj = 2 3 4 =1 2 +3
4 5 3 5 3 4
3 4 5
= 1( 1) 2( 2) + 3( 1) = 0 Singular J

(b)

2:11 0:80 1:72
jAj = 1:84 3:03 1:29
1:57 5:25 4:30
3:03 1:29 1:84 1:29 1:84 3:03
= 2:11 + 0:80 + 1:72
5:25 4:30 1:57 4:30 1:57 5:25
= 2:11(6:2565) + 0:80( 5: 8867) + 1:72( 4: 9029)
= 0:058 867 Ill conditioned J

(c)

2 1 0
2 1 1 1
jAj = 1 2 1 =2 +1
1 2 0 2
0 1 2
= 2(3) + 1( 2) = 4 Well-conditioned J

(d)

4 3 1
2 3 7 3 7 2
jAj = 7 2 3 =4 3 1
18 13 5 13 5 18
5 18 13
= 4(28) 3(76) 1( 116) = 0 Singular J




PROBLEM SET 2.1 1

,Problem 2

(a)
2 32 3 2 3
1 0 0 1 2 4 1 2 4
A = LU = 4 1 1 0 5 4 0 3 21 5 = 4 1 5 25 5 J
1 5=3 1 0 0 0 1 7 39
jAj = jLj jUj = (1 1 1)(1 3 0) = 0 J

(b)
2 32 3 2 3
2 0 0 2 1 1 4 2 2
A = LU = 4 1 1 0 5 4 0 1 3 5=4 2 2 4 5 J
1 3 1 0 0 1 2 4 11
jAj = jLj jUj = (2 1 1)(2 1 1) = 4 J



Problem 3

First solve Ly = b:
2 32 3 2 3
1 0 0 y1 1
4 3=2 1 0 5 4 y2 5 = 4 1 5
1=2 11=13 1 y3 2

y1 = 1
3 5
(1) + y2 = 1 y2 =
2 2
1 11 5 47
(1) + + y3 = 2 y3 =
2 13 2 13
Then solve Ux = y:
2 32 3 2 3
2 3 1 x1 1
4 0 13=2 7=2 5 4 x2 5 = 4 5=2 5
0 0 32=13 x3 47=13

32 47 47
x3 = x3 = J
13 13 32
13 7 47 5 13
x2 = x2 = J
2 2 32 2 32
13 47 59
2x1 3 = 1 x1 = J
32 32 32

2 PROBLEM SET 2.1

, Problem 4

The augmented coe¢cient matrix is
2 3
2 3 1 3
4
[Ajb] = 3 2 5 9 5
2 4 1 5
Elimination phase:

3
row 2 row 2 row 1
2
row 3 row 3 row 1
2 3
2 3 1 3
4 0 13=2 7=2 27=2 5
0 7 0 8

14
row 3 row 3 row 2
13
2 3
2 3 1 3
4 0 13=2 7=2 27=2 5
0 0 49=13 85=13
Solution by back substitution:

49 85 85
x3 = x3 = = 1:7347 J
13 13 49
13 7 85 27 8
x2 = x2 = = 1: 1429 J
2 2 49 2 7
8 85 32
2x1 3 = 3 x1 = = 0:6531 J
7 49 49



Problem 5

The augmented coe¢cient matrix is
2 3
2 0 1 0 1 0
6 0 1 2 0 0 0 7
[AjB] = 6
4
7
1 2 0 1 0 1 5
0 0 1 2 0 0

PROBLEM 4 3