Algorithm for Gauss elimination 1) first eliminate for each eq. j, j=1 to n-1
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Algorithm for Gauss elimination 1) first eliminate for each eq. j, j=1 to n-1 for all eq.s k greater than j a) multiply eq. j by a kj /a jj b) subtract the result from eq. K This leads to upper triangular
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Algorithm for Gauss elimination 1) first eliminate for each eq. j, j=1 to n-1 for all eq.s k greater than j a) multiply eq. j by a kj /a jj b) subtract the result from eq. K This leads to upper triangular. 2) now backsubstitute a) determine x n from b) put x n into n-1 eq. - PowerPoint PPT Presentation
Transcript of Algorithm for Gauss elimination 1) first eliminate for each eq. j, j=1 to n-1
Algorithm for Gauss elimination
1) first eliminate
for each eq. j, j=1 to n-1
for all eq.s k greater than j
a) multiply eq. j by akj/ajj
b) subtract the result from eq. K
This leads to upper triangular
2) now backsubstitute
a) determine xn from
b) put xn into n-1 eq.
c) solve for xn-1
d) repeat from b), moving back to n-2, n-3, etc. until all equations are solved
n
n
nn
n a
bx
1
1
Matlab code
Operation counting
Important as matrix gets large
For Gauss elimination
• elimination routine uses on the order ofoperations
• backsubstitution uses
3
3n
2
2nO
Problems with Gauss elimination (as done last time)
1) division by zero
2) round off errors
3) ill conditioned systems
Division by zero
1522
457
523
321
321
32
xxx
xxx
xx
Using first eq. to eliminate x1 in second eq. means dividing by 0
1522
42*0
153*
0
170*
0
11
523
321
321
32
xxx
xxx
xx
Pivoting developed to avoid this
find row with largest absolute value under pivot element
switch rows
More later
Round off errors
With more than n3/3 operations, get a lot of chopping.
More important - error is propagted
More than 100 equations - round can be very important - system dependent
Ill conditioned systems - small changes in coefficients lead to large changes in solution
Round-off errors especially important in ill-conditioned systems
Recall ill conditioned system from graphical methods
7.07.1
15.15.2
21
21
xx
xx
-2
-1
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6
x1
x2
2222121
1212111
bxaxa
bxaxa
Can write as
22
21
22
212
12
11
12
112
a
bx
a
ax
a
bx
a
ax
Since slopes are almost equal
22
21
12
11
a
a
a
a
22
21
12
11
a
a
a
a
21122211 aaaa becomes
021122211 aaaa
0det2221
1211
aa
aa
05.055.25.217.1
5.15.2det
Determinant close to zero indicates ill-conditioned set of equations.
How close?
No clear answer
Problem of scale
Multiply our set of equations by 100
70100170
100150250
21
21
xx
xx
500500,25000,25100170
150250det
-2
-1
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6
x1
x2
However, graphically
No change
Scaling
Make maximum element in any row =1
4118.05882.0
4.06.0
21
21
xx
xx70100170
100150250
21
21
xx
xx
0118.05882.01
6.01det
Another problem - singular systems
Two equations in the set are the same
Determinant is 0.
Calculating determinant using Gauss elimination
Given
nn
n
n
n
n
a
aa
aaa
aaaa
A
1
333
22322
1131211
''''
'''
nn
naaaaA 1332211 '''det
then
How to avoid pitfalls
•higher precision
•pivoting
•scaling
Partial Pivoting
Determine the largest coefficient in the column below pivot element
Then switch rows
(Compete pivot switches columns also, but is rarely used.)
Example:
4
11
3
7
653
3210
451
1273
Exchange rows 2 and 3
4
3
11
7
653
451
3210
1273
And now eliminate
Algorithm for Gauss elimination using improvements
1) first eliminate
for each eq. j, j=1 to n-1
first scale each equation k greater than jthen pivotnow
a) multiply eq. j by akj/ajj
b) subtract the result from eq. k
The Gauss Jordan method
Major difference - eliminate unknowns from all rows, not just subsequent ones
Normalize matrix so all entries are 1
Leads to identity matrix instead of upper triangular
Backsubstitution is easy
3
2
1
3
2
1
'
'
'
1
1
1
b
b
b
x
x
x
