Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

3.2. PIVOTING STRATEGIES FOR GAUSSIAN ELIMINATION 29
is maximized. The algorithm pivots on row l k , producing a bunch of zeros in the k th column.
The slick way to implement this is to first set l i = i for i = 1, 2, . . . , n. Then rearrange this
vector in a kind of “bubble sort”: when you find the index that should be l 1 , swap them, i.e., find
the j such that l j should be the first pivot and switch the values of l 1 , l j .
Then at the k th step, search only those indices in the tail of this vector: i.e., only among l j for
k ≤ j ≤ n, and perform a swap.
3.2.2 An Example
We present an example of using scaled partial pivoting with G.E. It’s hard to come up with an
example where the numbers do not come out as ugly fractions. We’ll look at a homework question.
⎛
⎜
⎝
2 −1 3 7 15
4 4 0 7 11
2 1 1 3 7
6 5 4 17 31
The scales are as follows: s 1 = 7, s 2 = 7, s 3 = 3, s 4 = 17.
We pick l 1 . It should be the index which maximizes |a i1 | /s i . These values are:
2
7 , 4 7 , 2 3 , 6 17 .
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⎟
⎠
We pick l 1 = 3, and pivot:
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⎜
⎝
0 −2 2 4 8
0 2 −2 1 −3
2 1 1 3 7
0 2 1 8 10
⎞
⎟
⎠
We pick l 2 . It should not be 3, and should be the index which maximizes |a i2 | /s i . These values
are:
2
7 , 2 7 , 2 17 .
We have a tie. In this case we pick the second row, i.e., l 2 = 2. We pivot:
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⎜
⎝
0 0 0 5 5
0 2 −2 1 −3
2 1 1 3 7
0 0 3 7 13
The matrix is in permuted upper triangular form. We could proceed, but would get a zero
multiplier, and no changes would occur.
If we did proceed we would have l 3 = 4. Then l 4 = 1. Our row permutation is 3, 2, 4, 1. When
we do back substitution, we work in this order reversed on the rows, solving x 4 , then x 3 , x 2 , x 1 .
We get x 4 = 1, so
x 3 = 1 (13 − 7 ∗ 1) = 2
3
⎞
⎟
⎠
x 2 = 1 (−3 − 1 ∗ 1 + 2 ∗ 2) = 0
2
x 1 = 1 (7 − 3 ∗ 1 − 1 ∗ 2 − 1 ∗ 0) = 1
2