Systems of Linear Algebraic Equations
Nizar Salim / Lecture
1.3.3 Gauss-Jordan Elimination
Gauss-Jordan elimination is a variation of Gauss elimination in
which the elements above the major diagonal are eliminated (made zero)
as well as the elements below the major diagonal. The A matrix is
transformed to a diagonal matrix. The rows are usually scaled to yield
unity diagonal elements, which transforms the A matrix to the identity
matrix, I. the transformed b vector is the solution vector x.
Example 9
Using Gauss-Jordan elimination, to find x1, x2, x3, in equation (1.2)
80x1 20x2 20x3 = 20
- 20x1 + 40x2 20x3 = 20
- 20x1 - 20x2 + 130x3 = 20
Solution
/ 80
20
20
20
20 20 130
20 40 20
80 20 20 1 R
Scaling row1 to give a11=1
3 1
2 1
( 20)
( 20)
20
20
1/ 4
20 20 130
20 40 20
1 1/ 4 1/ 4
R R
R R
/ 35
25
25
1/ 4
0 25 125
0 35 25
1 1/ 4 1/ 4
2 R
3 2
1 2
( 25)
( 1/ 4)
25
5 / 7
1/ 4
0 25 125
0 1 5 / 7
1 1/ 4 1/ 4
R R
R R
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
300/ 7 /(750/ 7)
5/ 7
3/ 7
0 0 750/ 7
0 1 5/ 7
1 0 3/ 7
3 R
2 3
1 3
( 5 / 7)
( 3/ 7)
2 / 5
5 / 7
3/ 7
0 0 1
0 1 5 / 7
1 0 3/ 7
R R
R R
0.40
1.00
0.60
0 0 1
0 1 0
1 0 0
The matrix has been transformed to the identity matrix I and the b vector
has been transformed to the solution vector, x
XT =[ 0.60 1.00 0.40 ]
1.4 Matrix inverse by Gauss-Jordan elimination
The inverse of a square matrix A is the matrix A-1 such that
A A-1 =A-1 A=I Gauss-Jordan elimination can be used to evaluate the
inverse of matrix A by augmenting A with the identity matrix I and
applying the Gauss-Jordan algorithm. The transformed matrix A is the
identity matrix I, and the transformed identity matrix is the matrix
inverse, A-1. thus, applying Gauss-Jordan elimination yields.
[A | I ] »» [I | A-1] ________________(1.23)
Example 10
Find the inverse of matrix A by using Gauss-Jordan elimination
Where
20 20 130
20 40 20
80 20 20
A
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
Solution
0 0 1
0 1 0
1 0 0
20 20 130
20 40 20
80 20 20
A
Performing Gauss-Jordan elimination
1/ 250 1/150 7 / 750
1/100 1/ 30 1/150
2/5 1/100 1/ 250
0 0 1
0 1 0
1 0 0
0.004000 0.006667 0.009333
0.010000 0.033333 0.006667
0.016000 0.010000 0.004000
1 / 250 1 / 150 7 / 750
1 / 100 1 / 30 1 / 150
2 / 125 1 / 100 1 / 250
A 1
The matrix inverse method
The system of linear algebraic equations can be solved by using the
matrix inverse, A-1 consider the general system of linear algebraic
equations:
Ax=b _____________(1.24)
Multiplying (1.24) by A-1 yields
AA-1x=A-1b
Ix=A-1b
x=A-1b _________________(1.25)
Thus, when the matrix inverse A-1 of the coefficient matrix A is
known, the solution vector x is simply the product of the matrix inverse
A-1 and the right-hand-side vector b.
Not all matrices have inverses singular matrices, that is matrices whose
determinant is zero, do not have inverses.
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
Example 11
Solve the linear system considered in example (10) using the matrix
inverse method.
Solution
20
20
20
1 / 250 1 / 150 7 / 750
1 / 100 1 / 30 1 / 150
2 / 125 1 / 100 1 / 250
x A 1b
x1=(2/125)(20) + (1/100)(20) + (1/250)(20)=0.60
x2=(1/100)(20) + (1/30)(20) + (1/150)(20)=1.00
x3=(1/250)(20) + (1/150)(20) + (7/750)(20)=0.04
Thus xT=[0.60 1.00 0.04]
1.5 Pitfalls of Elimination Method
All nonsingular system of linear algebraic equations have a solution.
In theory, the solution can always be obtained by Gauss elimination.
However, there are two major pitfalls in the application of Gauss
elimination (a)the presence of round-off errors, and (b)ill-condition
systems.
1.5.1 Round-off Errors
Round-off errors occur when exact infinite precision numbers are
approximated by finite precision numbers. The effects of round-off errors
are illustrated in the following example.
Example 12 Effects of round-off errors.
Consider the fallowing system of linear algebraic equation:
0.0003x1+3x2=1.0002 -------------------------------(1.26)
x1+x2 =1
Solve this equation by Gauss elimination, thus
1 / 0.0003
1.0002
1 1
0.0003 3
2 1 R R
-----------------------------(1.27)
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
0.0003
1.0002
1
1.0002
0 9999
0.0003 3
-----------(1.28)
3
1
9999
0.0003
0.9999
9999
0.0003
0.0003 1.0002
9999
0.0003
1.0002
1
2 x ---------(1.29)
3
2
0.0003
0.0002
0.0003
1.0002 3(1/3)
0.0003
1.0002 3 2
1
x
x ----------(1.30)
9999
0.0003
1.0002
1
2 x
0.0003
1.0002 3 2
1
x
x ------------(1.31)
The results are presented in table 1.1
Table 1.1 Solution of eq. (1.31)
Precision x2 x1
3 0.333 3.33
4 0.3333 1.333
5 0.33333 0.70000
6 0.333333 0.670000
7 0.3333333 0.6670000
8 0.33333333 0.66670000
Lets rework the problem by interchanging rows 1 and 2 in eq.(1.26)
x1+x2 =1
0.0003x1+3x2=1.0002 -------------------------------(1.32)
By Gauss elimination
2 1 1.0002 0.0003
1
0.0003 3
1 1
R R
------------------------(1.33)
0.9999
1
0 2.9997
1 1
-------------------------(1.34)
2 1 2 1
2.9997
0.9999
x and x x -------------------------(1.35)
Systems of Linear Algebraic Equations
Nizar Salim / Lecture
Lets solve eq.(1.35) using finite arithmetic. The results are presented in
Table 1.2
Table 1.2 Solution of equation (1.35)
precision x2 x1
3 0.333 0.667
4 0.3333 0.6667
5 0.33333 0.66667
This document was created with Win2PDF available at http://www.daneprairie.com.
The unregistered version of Win2PDF is for evaluation or non-commercial use only.
المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .