CHAPTER 20
Numeric Linear Algebra
SECTION 20.1. Linear Systems: Gauss Elimination, page 833
Purpose.
To explain the Gauss elimination, which is a solution method for linear systems
of equations by systematic elimination (reduction to triangular form).
Main Content, Important Concepts
Gauss elimination, back substitution
Pivot equation, pivot, choice of pivot
Operations count, order [e.g.,
O
(
n
3
)]
Comments on Content
This section is independent of Chap. 7 on matrices (in particular, independent of Sec. 7.3,
where the Gauss elimination is also considered).
Gauss’s method and its variants (Sec. 20.2) are the most important solution methods
for those systems (with matrices that do not have too many zeros).
The Gauss–Jordan method (Sec. 20.2) is less practical because it requires more
operations than the Gauss elimination.
Cramer’s rule (Sec. 7.7) would be totally impractical in numeric work, even for systems
of modest size.
SOLUTIONS TO PROBLEM SET 20.1, page 839
2.
x
1
5
0.65
x
2
,
x
2
arbitrary. Both equations represent the same straight line.
4.
x
1
5
0,
x
2
52
3
6.
x
1
5
(30.6
1
15.48
x
2
)/25.38,
x
2
arbitrary
8.
No solution; the matrix obtained at the end is
YZ
.
10.
x
1
5
0.5,
x
2
0.5,
x
3
5
3.5
12.
x
1
5
0.142857,
x
2
5
0.692308,
x
3
0.173913
14.
x
1
5
1.05,
x
2
5
0,
x
3
0.45,
x
4
5
0.5
16. Team Project.
(a) (i)
a
Þ
1 to make
D
5
a
2
1
Þ
0; (ii)
a
5
1,
b
5
3;
(iii)
a
5
1,
b
Þ
3.
(b)
x
1
5
1
_
2
(3
x
3
2
1),
x
2
5
1
_
2
(
2
5
x
3
1
7),
x
3
arbitrary is the solution of the first system.
The second system has no solution.
(c) det
A
5
0 can change to det
A
Þ
0 because of roundoff.
(d) (1
2
1/
e
)
x
2
5
2
2
1/
eventually becomes
x
2
/
<
1/
,
x
2
5
1,
x
1
5
(1
2
x
2
)/
<
0. The exact solution is
x
1
5
1/(1
2
),
x
2
5
(1
2
2
)/(1
2
).
We obtain it if we take
x
1
1
x
2
5
2 as the pivot equation.
(e) The exact solution is
x
1
5
1,
x
2
4. The 3-digit calculation gives
x
2
4.5,
x
1
5
1.27 without pivoting and
x
2
6,
x
1
5
2.08 with pivoting. This shows that
2
2
3
5
1
8
0
3
2
4
0
5
0
0
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1:21 PM
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