CS 257
Numerical Methods  Homework 4
October 5, 2006
1.
[1pt]
Section 7.1 #1
Solution:
A
=
1
4
α
2

1
2
α
α
3
1
b
=
6
3
5
det(
A
) = 9(
α

1)(
α
+ 1)
•
α
= 0: The determinant of
A
is nonzero, hence there exists a unique solution.
•
α
= 1: The determent of
A
is zero. To see if there are infinitely many solutions or no solutions, we use
Gaussian Elimination to create an equivalent system.
˜
A
=
1
0
1
0
1
0
0
0
0
˜
b
=
2
1
0
We effectively have two equations and three unknowns. Note that
x
2
= 1. We are free to pick a value
for
x
3
and find a value for
x
1
. There are infinitely many solutions.
•
α
=

1: The determinant of
A
is again zero and we create an equivalent system.
˜
A
=
1
0

1
0
1
0
0
0
0
˜
b
=
0
0
1
The last equation implies 0 = 1, which is false. This system has no solution.
If we replace the RHS by
b
= [0 0 0]
T
, then setting all three variables to 0 always solves the system.
•
α
= 0: The determinant of
A
is nonzero. The solution we found is unique.
•
α
= 1: Because the determinant is zero AND there exists one solution, we know there are infinitely many
solutions.
•
α
=

1: By the same reasoning as the above point, there are infinitely many solutions.
2.
[1pt]
Section 7.1 #4
Solution:
I will use rounding in my solution.
I will show the augmented matrix after each elimination.
All binary
operations are performed only to the given accuracy.
•
Four digits of precision:
0
.
1036
0
.
2122
0
.
7381
0
.
2081
0
.
4247
0
.
9327
0
.
1036
0
.
2122
0
.
7381
0

0
.
001600

0
.
5503
In the back substitution, we find
x
2
=

0
.
5503
0
.
001600
= 343
.
9
x
1
=
0
.
7381

0
.
2122
*
x
2
0
.
1036
=
0
.
7381

72
.
98
0
.
1036
=

72
.
24
0
.
1036
=

697
.
3
1
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•
Eight digits of precision:
0
.
1036
0
.
2122
0
.
7381
0
.
2081
0
.
4247
0
.
9327
0
.
1036
0
.
2122
0
.
7381
0

0
.
00154345

0
.
54991210
In the back substitution, we find
x
2
=

0
.
54991210

0
.
00154345
= 356
.
28760
x
1
=
0
.
7381

0
.
2122
*
x
2
0
.
1036
=
0
.
7381

75
.
604229
0
.
1036
=

74
.
866129
0
.
1036
=

722
.
64603
3.
[1pt]
Section 7.1 #7a
Solution:
3
4
3
10
1
5

1
7
6
3
7
15
3
4
3
10
0
3
.
667

2
3
.
667
0

5
1

5
3
4
3
10
0
3
.
667

2
3
.
667
0
0

1
.
728
.
001788
Performing the back substitution
x
3
=
.
001788

1
.
728
=

0
.
001035
x
2
=
3
.
667+2
x
3
3
.
667
=
3
.
667

0
.
00207
3
.
667
=
3
.
665
3
.
667
= 0
.
9995
x
1
=
10

(4
x
2+3
x
3
)
3
=
10

(3
.
998

0
.
003105)
3
=
10

3
.
995
3
=
6
.
005
3
= 2
.
002
4.
[1pt]
Section 7.2 #2
Solution:
The scale vector is [2 2 4].
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 Fall '05
 ThomasKerkhoven
 Addition, Pivot element

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