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04/21/23 http://numericalmethods.eng.usf.edu
1
Simultaneous Linear Equations
Topic: Gaussian Elimination
Gaussian Elimination
One of the most popular techniques for solving simultaneous linear equations of the form
Consists of 2 steps
1. Forward Elimination of Unknowns.
2. Back Substitution
CXA
Forward Elimination
The goal of Forward Elimination is to transform the coefficient matrix into an Upper Triangular Matrix
7.000
56.18.40
1525
112144
1864
1525
Forward Elimination
Linear Equations
A set of n equations and n unknowns
11313212111 ... bxaxaxaxa nn
22323222121 ... bxaxaxaxa nn
nnnnnnn bxaxaxaxa ...332211
. . . . . .
Forward Elimination
Transform to an Upper Triangular Matrix
Step 1: Eliminate x1 in 2nd equation using equation 1 as the pivot equation
)(1
2111
aa
Eqn
Which will yield
111
211
11
21212
11
21121 ... b
a
axa
a
axa
a
axa nn
Forward Elimination
Zeroing out the coefficient of x1 in the 2nd equation.
Subtract this equation from 2nd equation
111
2121
11
212212
11
2122 ... b
a
abxa
a
aaxa
a
aa nnn
'2
'22
'22 ... bxaxa nn
nnn aa
aaa
aa
aaa
111
212
'2
1211
2122
'22
Or Where
Forward Elimination
Repeat this procedure for the remaining equations to reduce the set of equations as
11313212111 ... bxaxaxaxa nn '2
'23
'232
'22 ... bxaxaxa nn
'3
'33
'332
'32 ... bxaxaxa nn
''3
'32
'2 ... nnnnnn bxaxaxa
. . . . . . . . .
Forward Elimination
Step 2: Eliminate x2 in the 3rd equation.
Equivalent to eliminating x1 in the 2nd equation using equation 2 as the pivot equation.
)(2
3 3222
aa
EqnEqn
Forward Elimination
This procedure is repeated for the remaining equations to reduce the set of equations as
11313212111 ... bxaxaxaxa nn '2
'23
'232
'22 ... bxaxaxa nn
"3
"33
"33 ... bxaxa nn
""3
"3 ... nnnnn bxaxa
. . . . . .
Forward Elimination
Continue this procedure by using the third equation as the pivot equation and so on.
At the end of (n-1) Forward Elimination steps, the system of equations will look like:
'2
'23
'232
'22 ... bxaxaxa nn
"3
"3
"33 ... bxaxa nn
11 nnn
nnn bxa
. . . . . .
11313212111 ... bxaxaxaxa nn
Forward Elimination
At the end of the Forward Elimination steps
)-(nnn
3
2
1
nnn
n
n
n
b
b
b
b
x
x
x
x
a
aa
aaa
aaaa
1
"3
'2
1
)1(
"3
"33
'2
'23
'22
1131211
Back Substitution
The goal of Back Substitution is to solve each of the equations using the upper triangular matrix.
3
2
1
3
2
1
33
2322
131211
x
x
x
00
0
b
b
b
a
aa
aaa
Example of a system of 3 equations
Back Substitution
Start with the last equation because it has only one unknown
)1(
)1(
n
nn
nn
n a
bx
Solve the second from last equation (n-1)th using xn solved for previously.
This solves for xn-1.
Back Substitution
Representing Back Substitution for all equations by formula
1
1
11
iii
n
ijj
iij
ii
i a
xabx For i=n-1, n-2,….,1
and
)1(
)1(
n
nn
nn
n a
bx
Example: Rocket Velocity
The upward velocity of a rocket is given at three different times
Time, t Velocity, v
s m/s
5 106.8
8 177.2
12 279.2
The velocity data is approximated by a polynomial as:
12.t5 , 322
1 atatatv
Find: The Velocity at t=6,7.5,9, and 11 seconds.
Example: Rocket Velocity
Assume
12.t5 ,atatatv 322
1
3
2
1
323
222
121
1
1
1
v
v
v
a
a
a
tt
tt
tt
3
2
1
Results in a matrix template of the form:
Using date from the time / velocity table, the matrix becomes:
2.279
2.177
8.106
112144
1864
1525
3
2
1
a
a
a
Example: Rocket Velocity
Forward Elimination: Step 1
)64(
25
12
RowRow
Yields
2.279
21.96
81.106
a
a
a
112144
56.18.40
1525
3
2
1
Example: Rocket Velocity
)144(
25
13
RowRow
0.336
21.96
8.106
a
a
a
76.48.160
56.18.40
1525
3
2
1
Yields
Forward Elimination: Step 1
Example: Rocket Velocity
Yields
)8.16(8.4
23
RowRow
735.0
21.96
8.106
a
a
a
7.000
56.18.40
1525
3
2
1
This is now ready for Back Substitution
Forward Elimination: Step 2
Example: Rocket Velocity
Back Substitution: Solve for a3 using the third equation
735.07.0 3 a
70
7350
.
.a 3
0501. a 3
Example: Rocket Velocity
Back Substitution: Solve for a2 using the second equation
21.9656.18.4 32 aa
8.4
56.121.96 32
aa
84
05015612196
.-
...-a 2
7019.a 2
Example: Rocket Velocity
Back Substitution: Solve for a1 using the first equation
8.106525 321 aaa
25
58.106 32
1
aaa
25
050.170.1958.1061
a
2900.01 a
Example: Rocket Velocity
Solution:The solution vector is
050.1
70.19
2900.0
3
2
1
a
a
a
The polynomial that passes through the three data points is then:
322
1 atatatv
125 ,050.170.192900.0 2 ttt
Example: Rocket Velocity
Solution:
Substitute each value of t to find the corresponding velocity
.s/m1.165
050.15.770.195.72900.05.7v 2
.s/m8.201
050.1970.1992900.09v 2
.s/m8.252
050.11170.19112900.011v 2
./ 69.129
050.1670.1962900.06 2
sm
v
Pitfalls
Two Potential Pitfalls-Division by zero: May occur in the forward elimination steps. Consider the set of equations:
655
901.33099.26
7710
321
321
32
xxx
xxx
xx
- Round-off error: Prone to round-off errors.
Pitfalls: Example
Consider the system of equations:
Use five significant figures with chopping
515
6099.23
0710
3
2
1
x
x
x
6
901.3
7
=
At the end of Forward Elimination
1500500
6001.00
0710
3
2
1
x
x
x
15004
001.6
7
=
Pitfalls: Example
Back Substitution
99993.015005
150043 x
5.1 001.0
6001.6 32
x
x
3500.010
077 321
xxx
15004
001.6
7
1500500
6001.00
0710
3
2
1
x
x
x
Pitfalls: Example
Compare the calculated values with the exact solution
99993.0
5.1
35.0
3
2
1
x
x
x
X calculated
1
1
0
3
2
1
x
x
x
X exact
Improvements
Increase the number of significant digits
Decreases round off error
Does not avoid division by zero
Gaussian Elimination with Partial Pivoting
Avoids division by zero
Reduces round off error
Partial Pivoting
pka
Gaussian Elimination with partial pivoting applies row switching to normal Gaussian Elimination.
How?
At the beginning of the kth step of forward elimination, find the maximum of
nkkkkk aaa .......,,........., ,1
If the maximum of the values is In the pth row, ,npk
then switch rows p and k.
Partial Pivoting
What does it Mean?
Gaussian Elimination with Partial Pivoting ensures that each step of Forward Elimination is performed with the pivoting element |akk| having the largest absolute value.
Partial Pivoting: Example
Consider the system of equations
6x5xx5
901.3x3x099.2x3
7x7x10
321
321
21
In matrix form
515
6099.23
0710
3
2
1
x
x
x
6
901.3
7
=
Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping
Partial Pivoting: Example
Forward Elimination: Step 1
Examining the values of the first column
|10|, |-3|, and |5| or 10, 3, and 5
The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1.
6
901.3
7
515
6099.23
0710
3
2
1
x
x
x
5.2
001.6
7
55.20
6001.00
0710
3
2
1
x
x
x
Performing Forward Elimination
Partial Pivoting: Example
Forward Elimination: Step 2
Examining the values of the first column
|-0.001| and |2.5| or 0.0001 and 2.5
The largest absolute value is 2.5, so row 2 is switched with row 3
5.2
001.6
7
55.20
6001.00
0710
3
2
1
x
x
x
001.6
5.2
7
6001.00
55.20
0710
3
2
1
x
x
x
Performing the row swap
Partial Pivoting: Example
Forward Elimination: Step 2
Performing the Forward Elimination results in:
002.6
5.2
7
002.600
55.20
0710
3
2
1
x
x
x
Partial Pivoting: Example
Back Substitution
Solving the equations through back substitution
1002.6
002.63 x
15.2
55.2 22
xx
010
077 321
xxx
002.6
5.2
7
002.600
55.20
0710
3
2
1
x
x
x
Partial Pivoting: Example
1
1
0
3
2
1
x
x
x
X exact
1
1
0
3
2
1
x
x
x
X calculated
Compare the calculated and exact solution
The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting
Summary
-Forward Elimination
-Back Substitution
-Pitfalls
-Improvements
-Partial Pivoting
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