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Simultaneous Linear Equations
Topic: Gaussian Elimination
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Gaussian Elimination: Solving Linear Equations with Elimination and Substitution, Lecture notes of Differential Equations
The Gaussian Elimination method for solving simultaneous linear equations. It covers the steps of Forward Elimination and Back Substitution, as well as common pitfalls and improvements such as partial pivoting.
Typology: Lecture notes
2019/2020
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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
- Forward Elimination of Unknowns.
- Back Substitution
A X C
Forward Elimination
Linear Equations
A set of n equations and n unknowns
11 1 12 2 13 3 1 1
a x a x a x ... a x b
n n
21 1 22 2 23 3 2 2
a x a x a x ... a x b
n n
n n n nn n n
a x a x a x ... a x b
1 1 2 2 3 3
.. .. ..
Forward Elimination
Transform to an Upper Triangular Matrix
Step 1: Eliminate x 1 in 2
nd equation using equation 1 as
the pivot equation
( )
1
21
11
a
a
Eqn
Which will yield
1
11
21
1
11
21
12 2
11
21
21 1
... b
a
a
a x
a
a
a x
a
a
a x
n n
Forward Elimination
Repeat this procedure for the remaining
equations to reduce the set of equations as
11 1 12 2 13 3 1 1
a x a x a x ... a x b
n n
'
2
'
3 2
'
2 23
'
22
a x a x ... a x b
n n
'
3
'
3 3
'
2 33
'
32
a x a x ... a x b
n n
' '
3
'
2 3
'
2
... n n nn n n
a x a x a x b
... ... ...
Forward Elimination
Step 2: Eliminate x 2
in the 3
rd equation.
Equivalent to eliminating x 1 in the 2
nd equation
using equation 2 as the pivot equation.
32
22
a
a
Eqn
Eqn
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
'
3 2
'
2 23
'
22
a x a x ... a x b
n n
"
3
"
3
"
33
a x ... a x b n n
1 ^ ^1
n
n n
n
nn
a x b
.. .. ..
11 1 12 2 13 3 1 1
a x a x a x ... a x b
n n
Forward Elimination
At the end of the Forward Elimination steps
(n- )
n n
3
2
1
n
nn
n
n
n
b
b
b
b
x
x
x
x
a
a a
a a a
a a a a
1
"
3
'
2
1
( 1 )
"
3
"
33
'
2
'
23
'
22
11 12 13 1
Back Substitution
Start with the last equation because it has only
one unknown
( 1 )
( 1 )
n
nn
n
n
n
a
b
x
Solve the second from last equation (n-1)
th
using x n
solved for previously.
This solves for x n-
.
Back Substitution
Representing Back Substitution for all equations
by formula
1
1
1 1
i
ii
n
j i
j
i
ij
i
i
i
a
b a x
x
For i = n -1, n -2,….,
and
( 1 )
( 1 )
n
nn
n
n
n
a
b
x
Example: Rocket Velocity
Assume
v^ ^ t ^ a t a t a , 5 t 12.
2 3
2 1
3
2
1
3
2
3
2
2 2
1
2
1
1
1
1
v
v
v
a
a
a
t t
t t
t t
3
2
1
Results in a matrix template of the form:
Using date from the time / velocity table, the matrix becomes:
2
2
8
144 12 1
64 8 1
25 5 1
3
2
1
a
a
a
Example: Rocket Velocity
Forward Elimination: Step 1
Row
Row
Yields
a
a
a
3
2
1
Example: Rocket Velocity
Yields
Row
Row
a
a
a
3
2
1
This is now ready for Back Substitution
Forward Elimination: Step 2
Example: Rocket Velocity
Back Substitution: Solve for a 3
using the third equation
3
a
a
3