Page 25 - Matrix Analysis & Applied Linear Algebra

1.3 Gauss–Jordan Method 17

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number of additions/subtractions. Compare this with the n /2 factor required
by the Gauss–Jordan method, and you can see that Gauss–Jordan requires about
50% more effort than Gaussian elimination with back substitution. For small sys-
tems of the textbook variety (e.g., n = 3 ), these comparisons do not show a great
deal of difference. However, in practical work, the systems that are encountered
can be quite large, and the difference between Gauss–Jordan and Gaussian elim-
ination with back substitution can be significant. For example, if n = 100, then
3
3
n /3 is about 333,333, while n /2 is 500,000, which is a difference of 166,667
multiplications/divisions as well as that many additions/subtractions.
Although the Gauss–Jordan method is not recommended for solving linear
systems that arise in practical applications, it does have some theoretical advan-
tages. Furthermore, it can be a useful technique for tasks other than computing
solutions to linear systems. We will make use of the Gauss–Jordan procedure
when matrix inversion is discussed—this is the primary reason for introducing
Gauss–Jordan.
Exercises for section 1.3

1.3.1. Use the Gauss–Jordan method to solve the following system:

4x 2 − 3x 3 =3,
−x 1 +7x 2 − 5x 3 =4,
−x 1 +8x 2 − 6x 3 =5.

1.3.2. Apply the Gauss–Jordan method to the following system:

x 1 + x 2 + x 3 + x 4 =1,
x 1 +2x 2 +2x 3 +2x 4 =0,
x 1 +2x 2 +3x 3 +3x 4 =0,
x 1 +2x 2 +3x 3 +4x 4 =0.

1.3.3. Use the Gauss–Jordan method to solve the following three systems at
the same time.
2x 1 − x 2 =1 0 0,
−x 1 +2x 2 − x 3 =0 1 0,
−x 2 + x 3 =0 0 1.

1.3.4. Verify that the operation counts given in the text for the Gauss–Jordan
method are correct for a general 3 × 3 system. If you are up to the
challenge, try to verify these counts for a general n × n system.