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Simultaneous Linear Equations
Topic: Gaussian Elimination
Dr. Nasir M Mirza
Numerical Methods
Email: [email protected]
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Gaussian Elimination
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Basic Principles
The general description of a set of linear equations in the matrixform: [A][X] = [B]
Where, [A] is ( m x n ) matrix, the [X] is a ( n x 1 ) vector, andthe [B] is a (m x 1 ) vector.
How we solve such a system:
Write the equations in natural form
Identify unknowns and order them
Isolate unknowns
Write equations in matrix form
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Types of Matrix Formulation
If m = n The solution of [A]{x} ={b} with n unknowns and m(n) equations
If m > n The system is overdetermined system (Least Square Problems)
If m < n The system is underdetermined system (Optimization Problems)
mnmnmmm
nn
nn
nn
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
332211
33333232131
22323222121
11313212111
Suppose we have an ( m x n ) Array
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Matrix Representation
nnnnnnn
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaa
aaaa
3
2
1
3
2
1
321
3333231
2232221
1131211
mnmnmmm
nn
nn
nn
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
332211
33333232131
22323222121
11313212111The set of n linear equationswith n unknowns:
The matix form:
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Consistency
nnnnnnn
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaaaaaa
bXA
3
2
1
3
2
1
321
3333231
2232221
1131211
The problem is consistent, if a solution exists for the problem.
The problem is inconsistent, if there is no solution for the problem.
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Rank of Matrix
Rank of a matrix is the number of linearlyindependent column vectors in the matrix.
For n x n matrix, A:
If rank(A) = n and is consistent, A has an uniquesolution exists
If rank(A) = n and is inconsistent, A has no solution
exists If rank(A) < n and system is consistent, A has an
infinite number of solutions
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Matrix
For an n x n system, rank(A) = n automaticallyguarantees
that the system is consistent.
The columns of A are linearly independent The rows of A are linearly independent
rank(A) = n
det(A) ~= 0
A-1 exists;
The solution to [A]{x} ={b} exist and is unique.
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Matrix Definition
1
6
15.0
12
y
x
5
6
12
12
y
x
Considery = -2.0 x + 6
y = 0.5 x + 1
There are 2 unknowns x , y
and rank is 2
Considery = -2 x + 6
y = -2 x + 5
There are 2 unknowns x , y and
rank is 1 and is inconsistent
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Gaussian Elimination
Gaussian Elimination is one of the most popular techniques for
solving simultaneous linear equations of the form: [A][X] =[b]
The method consists of 2 steps
1. Forward Elimination of Unknowns.2. Back Substitution
Let us learn the method first by examples:
mnmnmmm
nn
nn
nn
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
332211
33333232131
22323222121
11313212111
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Example 1:Consider following three linear equations with three unknowns:
)3(74.506.113.569.3
)2(69.475.378.046.1
)1(76.128.406.337.2
321
321
321
xxx
xxx
xxx
)4(7426.08059.12911.1 321 xxx
)5(6058.3366.66650.20
69.475.3780.046.1
0842.16366.2885.146.1
32
321
321
xx
xxx
xxx
First step is to divide by 2.37 so that the coefficient of x1 is one. Thus
Now multiply this by -1.46 and adding it with Eq2.
)6(4802.86038.58942.90 32 xx
Now multiply Eq.4 with 3.69 and adding it to Eq3.
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Example 1:Re-writing Equations 4, 5 and 6:
)6(4802.86038.58942.9)0(
)5(6058.33866.66650.2)0(
)4(7426.08059.12911.1
321
321
321
xxx
xxx
xxx
)7(3530.13965.20 32 xx
8671.211077.18)0(0 32 xx
The second step is to divide by -2.6650 so that the coefficient of x2 is one. Thus
Now multiply this by -9.8942 and adding it to Eq 6.
dividing it by 18.1077, we get
)8(2076.1)0(0 32 xx
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Example 1:Final form of Eq. 4, 7 and 8 is
)8(2076.100
)7(3530.13965.20
)4(7426.08059.12911.1
3
32
321
x
xx
xxx
5410.12076.13965.23530.13965.23530.1 32
xx
Now x3 is known and we can back-substitute it into Eq. 7 to find x2.
Similarly, we can find x1 using values of x2 and x3 from Eq. 4
Therefore, the solution is given as
x1= 0.9338 ; x
2= 1.5410 ; x
3= 1.2076
9338.0
2076.13965.25410.12911.17426.0
3965.22911.17426.0321
xxx
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Forward Elimination
The goal of Forward Elimination is to transform the coefficientmatrix into an Upper Triangular Matrix
7.000
56.18.401525
112144
1864
1525
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Forward Elimination
Linear Equations:A set ofn equations and n unknowns
11313212111 ... bxaxaxaxa nn
22323222121 ... bxaxaxaxa nn
nnnnnnn bxaxaxaxa ...332211
. .
. .
. .
( Eq.1 )
( Eq.2 )
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Forward Elimination
Transform to an Upper Triangular Matrix
Step 1: Eliminate x1 in 2nd equation using equation 1 as the pivot equation
)(1
21
11
aa
Eqn
Which will change the Eq. 1 as following:
1
11
21
1
11
21
212
11
21
121 ... ba
axa
a
axa
a
axa nn
11313212111 ... bxaxaxaxa nn (Eq.1)
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Forward Elimination
Zeroing out the coefficient of x1
in the 2nd equation.
Subtract Equation 1 from Eq.2
1
11
2121
11
212212
11
2122 ... b
a
abxa
a
aaxa
a
aa nnn
'
2
'
22
'
22 ... bxaxa nn
nnn aa
aaa
aaaaa
1
11
212
'
2
12
11
2122
'22
Or
1
11
21
1
11
21
212
11
21
121... b
a
axa
a
axa
a
axa
nn
Where
22323222121 ... bxaxaxaxa nn ( Eq.2 )
( Eq.1 )
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Forward Elimination
Repeat this procedure for the remaining equations to reduce the set ofequations as
11313212111 ... bxaxaxaxa nn '
2
'
23
'
232
'
22 ... bxaxaxa nn
'
3
'
33
'
332
'
32 ... bxaxaxa nn
''
3
'
32
'
2 ... nnnnnn bxaxaxa
. . .
. . .
. . .
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Forward Elimination
Step 2: Eliminate x2 in the 3rd equation.Equivalent to eliminating x1 in the 2
nd equation using equation 2 as the pivot
equation.
)(23 3222
aa
EqnEqn
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Forward Elimination
This procedure is repeated for the remaining equations to reduce theset of equations as
11313212111 ... bxaxaxaxa nn
'2
'23
'232
'22 ... bxaxaxa nn
"
3
"
33
"
33 ... bxaxa nn
""
3
"
3 ... nnnnn bxaxa
. .
. .
. .
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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 n
nn
n
nn bxa
. .
. .
. .
11313212111 ... bxaxaxaxa nn
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Forward Elimination
At the end of the Forward Elimination steps:
)-(n
nn
3
2
1
n
nn
n
n
n
b
bb
b
x
xx
x
a
aaaaa
aaaa
1
"
3
'
2
1
)1(
"
3
"
33
'
2
'
23
'
22
1131211
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Back Substitution
The goal of Back Substitution is to solve each of the equations usingthe 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
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Back Substitution
Start with the last equation because it has only
one unknown
)1(
)1(
n
nn
n
n
n
a
bx
Solve the second from last equation (n-1)th using xn solved for
previously.
This solves for xn-1.
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Back Substitution
Representing Back Substitution for all equations by formula
1
1
11
iii
n
ijj
i
ij
i
i
i
a
xab
x
Fori=n-1,
n-2,.,1
and)1(
)1(
n
nn
n
n
n
a
bx
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Computer Programfunction x = gaussE(A,b,ptol)
% GEdemo Show steps in Gauss elimination and back substitution% No pivoting is used.
%
% Synopsis: x = GEdemo(A,b)
% x = GEdemo(A,b,ptol)
%
% Input: A,b = coefficient matrix and right hand side vector
% ptol = (optional) tolerance for detection of zero pivot
% Default: ptol = 50 * eps
%
% Output: x = solution vector, if solution exists
A=[25 5 1; 64 8 1; 144 12 1]
b=[106.8; 177.2; 279.2]
if nargin
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Computer Program (continued)% program continued
for i =1:n-1pivot = Ab(i,i);
if abs(pivot)