Ch9-Gauss_Elimination4.pdf

The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Numerical Analysis
ECIV 3306
Chapter 9
Gauss Elimination
1

Part 3
Linear Algebraic Equations
• An equation of the form ax+by+c=0 or equivalently
ax+by=-c is called a linear equation in x and y variables.
• ax+by+cz=d is a linear equation in three variables, x, y,
and z.
• Thus, a linear equation in n variables is
a1x1+a2x2+ … +anxn = b
• A solution of such an equation consists of real numbers
c1, c2, c3, … , cn. If you need to work more than one
linear equations, a system of linear equations must be
solved simultaneously.
2

Noncomputer Methods for Solving
Systems of Equations
• For small number of equations (n 3) linear
equations can be solved readily by simple
techniques such as “method of elimination.”
• Linear algebra provides the tools to solve such
systems of linear equations.
• Nowadays, easy access to computers makes
the solution of large sets of linear algebraic
equations possible and practical.
3

Introduction
0
)
(x
f
Roots of a single equation:
A general set of
equations:
- n equations,
- n unknowns.
0
)
,
,
(
0
)
,
,
(
0
)
,
,
(
2
1
2
1
2
2
1
1
n
n
n
n
x
x
x
f
x
x
x
f
x
x
x
f
5

Linear Algebraic Equations
n
n
nn
n
n
n
n
x
n
n
b
x
a
x
a
x
a
b
x
x
a
e
a
x
a
b
x
a
x
x
a
x
a
2
2
1
1
2
3
2
2
22
3
1
21
1
5
1
2
1
12
1
11
/
)
(
)
(
)
(
2
n
n
nn
n
n
n
n
n
n
b
x
a
x
a
x
a
b
x
a
x
a
x
a
b
x
a
x
a
x
a
2
2
1
1
2
2
2
22
1
21
1
1
2
12
1
11
Nonlinear Equations
6

Review of Matrices
m
n
nm
2
n
1
n
m
2
22
21
m
1
12
11
a
a
a
a
a
a
a
a
a
]
A
[
2nd row
mth column
Elements are indicated by a i j
row column
Row vector: Column vector:
n
1
n
2
1 r
r
r
]
R
[
1
m
m
2
1
c
c
c
]
C
[
Square matrix:
- [A]nxm is a square matrix if n=m.
- A system of n equations with n unknonws has a square
coefficient matrix.
7

Review of Matrices
• Main (principle) diagonal:
[A]nxn consists of elements aii ; i=1,...,n
• Symmetric matrix:
If aij = aji [A]nxn is a symmetric matrix
• Diagonal matrix:
[A]nxn is diagonal if aij = 0 for all i=1,...,n ; j=1,...,n
and i j
• Identity matrix:
[A]nxn is an identity matrix if it is diagonal with aii=1
i=1,...,n . Shown as [I]
8

Review of Matrices
• Upper triangular matrix:
[A]nxn is upper triangular if aij=0 i=1,...,n ; j=1,...,n and i>j
• Lower triangular matrix:
[A]nxn is lower triangular if aij=0 i=1,...,n ; j=1,...,n and i<j
• Inverse of a matrix:
[A]-1 is the inverse of [A]nxn if [A]-1[A] = [I]
• Transpose of a matrix:
[B] is the transpose of [A]nxn if bij=aji Shown as [A] or [A]T
9

Special Types of Square Matrices
88
6
39
16
6
9
7
2
39
7
3
1
16
2
1
5
]
[A
nn
a
a
a
D 22
11
]
[
1
1
1
1
]
[ I
Symmetric Diagonal Identity
nn
n
n
a
a
a
a
a
a
A]
[ 2
22
1
12
11
nn
n a
a
a
a
a
A
1
22
21
11
]
[
Upper Triangular Lower Triangular
10

Review of Matrices
• Matrix multiplication:
r
1
k
kj
ik
ij b
a
c
Note: [A][B] [B][A]
11

Review of Matrices
• Augmented matrix: is a special way of showing two
matrices together.
For example augmented with the
column vector is
• Determinant of a matrix:
A single number. Determinant of [A] is shown as |A|.
22
21
12
11
a
a
a
a
A
2
1
b
b
B
2
22
21
1
12
11
b
a
a
b
a
a
12

Solving Small Numbers of Equations
There are many ways to solve a system of linear
equations:
• Graphical method
• Cramer’s rule
• Method of elimination
• Numerical methods for solving larger number of
linear equations:
- Gauss elimination (Chp.9)
- LU decompositions and matrix inversion(Chp.10)
For n 3
13
Gauss Elimination
Chapter 9

1. Graphical method
• For two equations (n = 2):
• Solve both equations for x2: the intersection of the lines
presents the solution.
• For n = 3, each equation will be a plane on a 3D coordinate
system. Solution is the point where these planes intersect.
• For n > 3, graphical solution is not practical.
2
2
22
1
21
1
2
12
1
11
b
x
a
x
a
b
x
a
x
a
22
2
1
22
21
2
1
2
12
1
1
12
11
2 intercept
(slope)
a
b
x
a
a
x
x
x
a
b
x
a
a
x
14

Graphical Method -Example
• Solve:
• Plot x2 vs. x1, the
intersection of the
lines presents the
solution.
2
2
18
2
3
2
1
2
1
x
x
x
x
15

Graphical Method
No solution Infinite solution ill condition
(sensitive to round-off errors)
16

2.Determinants and Cramer’s Rule
Determinant can be illustrated for a set of three
equations:
Where [A] is the coefficient matrix:
B
x
A .
33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
a
A
17

Cramer’s Rule
22
31
32
21
32
31
22
21
13
23
31
33
21
33
31
23
21
12
23
32
33
22
33
32
23
22
11
33
32
31
23
22
21
13
12
11
a
a
a
a
a
a
a
a
D
a
a
a
a
a
a
a
a
D
a
a
a
a
a
a
a
a
D
a
a
a
a
a
a
a
a
a
D
32
31
22
21
13
33
31
23
21
12
33
32
23
22
11
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
D
D
a
a
b
a
a
b
a
a
b
x
33
32
3
23
22
2
13
12
1
1
D
a
b
a
a
b
a
a
b
a
x
33
3
31
23
2
21
13
1
1
1
2
D
b
a
a
b
a
a
b
a
a
x 3
32
31
2
22
21
1
12
11
3
18

Cramer’s Rule
• For a singular system D = 0 Solution can not
be obtained.
• For large systems Cramer’s rule is not practical
because calculating determinants is costly.
• Example
19

Cramer’s Rule
• Example
20

3. Method of Elimination
• The basic strategy is to multiply the equations
by constants so that one of the unknowns will
be eliminated when the two equations are
combined. The result is a single equation that
can be solved for the remaining unknown.
• The elimination of unknowns can be extended to
systems with more than two or three equations.
However, the method becomes extremely
tedious to solve by hand.
21

Elimination of Unknowns Method
Given a 2x2 set of equations:
2.5x1 + 6.2x2 = 3.0
4.8x1 - 8.6x2 = 5.5
• Multiply the 1st eqn by 8.6
and the 2nd eqn by 6.2
21.50x1 + 53.32x2=25.8
29.76x1 – 53.32x2=34.1
• Add these equations 51.26 x1 + 0 x2 = 59.9
• Solve for x1 : x1 = 59.9/51.26 = 1.168552478
• Using the 1st eqn solve for x2 :
x2 =(3.0–2.5*1.168552478)/6.2 = 0.01268045242
• Check if these satisfy the 2nd eqn:
4.8*1.168552478–8.6*0.01268045242 = 5.500000004
(Difference is due to the round-off errors).
22

Naive Gauss Elimination Method
• It is a formalized way of the previous elimination
technique to large sets of equations by developing a
systematic scheme or algorithm to eliminate
unknowns and to back substitute.
• As in the case of the solution of two equations, the
technique for n equations consists of two phases:
1. Forward elimination of unknowns.
2. Back substitution.
23

• Consider the following system of n equations.
a11x1 + a12x2 + ... + a1nxn = b1 (1)
a21x1 + a22x2 + ... + a2nxn = b2 (2)
...
an1x1 + an2x2 + ... + annxn = bn (n)
Form the augmented matrix of [A|B].
Step 1 : Forward Elimination: Reduce the system to an upper triangular
system.
1.1- First eliminate x1 from 2nd to nth equations.
- Multiply the 1st eqn. by a21/a11 & subtract it from the 2nd equation.
This is the new 2nd eqn.
- Multiply the 1st eqn. by a31/a11 & subtract it from the 3rd equation.
This is the new 3rd eqn.
...
- Multiply the 1st eqn. by an1/a11 & subtract it from the nth equation.
This is the new nth eqn.
24

Note:
- In these steps the 1st eqn is the pivot equation and a11 is
the pivot element.
- Note that a division by zero may occur if the pivot
element is zero. Naive-Gauss Elimination does not check
for this.
The modified system is
‘ indicates that the
system is modified once.
n
3
2
1
n
3
2
1
nn
3
n
2
n
n
3
33
32
n
2
23
22
n
1
13
12
11
b
b
b
b
x
x
x
x
a
a
a
0
a
a
a
0
a
a
a
0
a
a
a
a
Naive Gauss Elimination Method (cont’d)
25

1.2- Now eliminate x2 from 3rd to nth equations.
The modified system is
n
3
2
1
n
3
2
1
nn
3
n
n
3
33
n
2
23
22
n
1
13
12
11
b
b
b
b
x
x
x
x
a
a
0
0
a
a
0
0
a
a
a
0
a
a
a
a
Repeat steps (1.1) and (1.2) upto (1.n-1).
11 12 13 1 1 1
22 23 2 2 2
33 3 3 3
( 1) ( 1)
0
0 0
0 0 0 0
n
n
n
n n
nn n n
a a a a x b
a a a x b
a a x b
a x b
we will get this upper
triangular system
26

Step 2 : Back substitution
Find the unknowns starting from the last equation.
1. Last equation involves only xn. Solve for it.
2. Use this xn in the (n-1)th equation and solve for xn-1.
...
3. Use all previously calculated x values in the 1st eqn
and solve for x1.
)
1
(
)
1
(
n
nn
n
n
n
a
b
x
1
2
1
for
)
1
(
1
1
)
1
(
, ...,
, n-
n-
i
a
x
a
b
x i
ii
n
i
j
j
i
ij
i
i
i
27

Summary of Naive Gauss Elimination Method
28

Example 1
Solve the following system using Naive Gauss Elimination.
6x1 – 2x2 + 2x3 + 4x4 = 16
12x1 – 8x2 + 6x3 + 10x4 = 26
3x1 – 13x2 + 9x3 + 3x4 = -19
-6x1 + 4x2 + x3 - 18x4 = -34
Step 0: Form the augmented matrix
6 –2 2 4 | 16
12 –8 6 10 | 26 R2-2R1
3 –13 9 3 | -19 R3-0.5R1
-6 4 1 -18 | -34 R4-(-R1)
29

Example 1 (cont’d)
Step 2: Back substitution
Find x4 x4 =(-3)/(-3) = 1
Find x3 x3 =(-9+5*1)/2 = -2
Find x2 x2 =(-6-2*(-2)-2*1)/(-4) = 1
Find x1 x1 =(16+2*1-2*(-2)-4*1)/6= 3
31

Naive Gauss Elimination Method Example 2
(Using 6 Significant Figures)
3.0 x1 - 0.1 x2 - 0.2 x3 = 7.85
0.1 x1 + 7.0 x2 - 0.3 x3 = -19.3 R2-(0.1/3)R1
0.3 x1 - 0.2 x2 + 10.0 x3 = 71.4 R3-(0.3/3)R1
Step 1: Forward elimination
3.00000 x1- 0.100000 x2 - 0.200000 x3 = 7.85000
7.00333 x2 - 0.293333 x3 = -19.5617
- 0.190000 x2 + 10.0200 x3 = 70.6150
3.00000 x1- 0.100000 x2 - 0.20000 x3 = 7.85000
7.00333 x2 - 0.293333 x3 = -19.5617
10.0120 x3 = 70.0843
32

Naive Gauss Elimination Method Example 2
(cont’d)
Step 2: Back substitution
x3 = 7.00003
x2 = -2.50000
x1 = 3.00000
Exact solution:
x3 = 7.0
x2 = -2.5
x1 = 3.0
33

Pitfalls of Gauss Elimination Methods
1. Division by zero
2 x2 + 3 x3 = 8
4 x1 + 6 x2 + 7 x3 = -3
2 x1 + x2 + 6 x3 = 5
It is possible that during both elimination and back-
substitution phases a division by zero can occur.
2. Round-off errors
In the previous example where up to 6 digits were kept
during the calculations and still we end up with close to
the real solution.
x3 = 7.00003, instead of x3 = 7.0
a11 = 0
(the pivot element)
34

Pitfalls of Gauss Elimination (cont’d)
3. Ill-conditioned systems
x1 + 2x2 = 10
1.1x1 + 2x2 = 10.4
x1 + 2x2 = 10
1.05x1 + 2x2 = 10.4
Ill conditioned systems are those where small changes in
coefficients result in large change in solution. Alternatively, it
happens when two or more equations are nearly identical,
resulting a wide ranges of answers to approximately satisfy the
equations. Since round off errors can induce small changes in the
coefficients, these changes can lead to large solution errors.
x1 = 4.0 & x2 = 3.0
x1 = 8.0 & x2 = 1.0
35

Pitfalls of Gauss Elimination (cont’d)
4. Singular systems.
• When two equations are identical, we would loose one
degree of freedom and be dealing with case of n-1
equations for n unknowns.
To check for singularity:
• After getting the forward elimination process and
getting the triangle system, then the determinant for
such a system is the product of all the diagonal
elements. If a zero diagonal element is created, the
determinant is Zero then we have a singular system.
• The determinant of a singular system is zero.
36

Techniques for Improving Solutions
1. Use of more significant figures to solve for the
round-off error.
2. Pivoting. If a pivot element is zero, elimination step
leads to division by zero. The same problem may arise,
when the pivot element is close to zero. This Problem
can be avoided by:
Partial pivoting. Switching the rows so that the
largest element is the pivot element.
Complete pivoting. Searching for the largest element
in all rows and columns then switching.
3. Scaling
Solve problem of ill-conditioned system.
Minimize round-off error
37

Partial Pivoting
Before each row is normalized, find the largest
available coefficient in the column below the
pivot element. The rows can then be switched
so that the largest element is the pivot
element so that the largest coefficient is used
as a pivot.
38

Use of more significant figures to solve for the
round-off error :Example.
Use Gauss Elimination to solve these 2 equations:
(keeping only 4 sig. figures)
0.0003 x1 + 3.0000 x2 = 2.0001
1.0000 x1 + 1.0000 x2 = 1.000
0.0003 x1 + 3.0000 x2 = 2.0001
- 9999.0 x2 = -6666.0
Solve: x2 = 0.6667 & x1 = 0.0
The exact solution is x2 = 2/3 & x1 = 1/3
39

Use of more significant figures to solve for the round-
off error :Example (cont’d).
3
2
2
x
0003
.
0
)
3
/
2
(
3
0001
.
2
1
x
Significant
Figures x2 x1
3 0.667 -3.33
4 0.6667 0.000
5 0.66667 0.3000
6 0.666667 0.33000
7 0.6666667 0.333000
40

Pivoting: Example
Now, solving the pervious example using the partial
pivoting technique:
1.0000 x1+ 1.0000 x2 = 1.000
0.0003 x1+ 3.0000 x2 = 2.0001
The pivot is 1.0
1.0000 x1+ 1.0000 x2 = 1.000
2.9997 x2 = 1.9998
x2 = 0.6667 & x1=0.3333
Checking the effect of the # of significant digits:
# of dig x2 x1
4 0.6667 0.3333
5 0.66667 0.33333
41

Scaling: Example
• Solve the following equations using naïve gauss elimination:
(keeping only 3 sig. figures)
2 x1+ 100,000 x2 = 100,000
x1 + x2 = 2.0
• Forward elimination:
2 x1+ 100,000 x2 = 100,000
- 50,000 x2 = -50,000
Solve x2 = 1.00& x1 = 0.00
• The exact solution is x1 = 1.00002 & x2 = 0.99998
42

Scaling: Example (cont’d)
B) Using the scaling algorithm to solve:
2 x1+ 100,000 x2 = 100,000
x1 + x2 = 2.0
Scaling the first equation by dividing by 100,000:
0.00002 x1+ x2 = 1.0
x1+ x2 = 2.0
Rows are pivoted:
x1 + x2 = 2.0
0.00002 x1+ x2 = 1.0
Forward elimination yield:
x1 + x2 = 2.0
x2 = 1.00
Solve: x2 = 1.00 & x1 = 1.00
The exact solution is x1 = 1.00002 & x2 = 0.99998 43

Scaling: Example (cont’d)
C) The scaled coefficient indicate that pivoting is necessary.
We therefore pivot but retain the original coefficient to give:
x1 + x2 = 2.0
2 x1+ 100,000 x2 = 100,000
Forward elimination yields:
x1 + x2 = 2.0
100,000 x2 = 100,000
Solve: x2 = 1.00 & x1 = 1.00
Thus, scaling was useful in determining whether pivoting was
necessary, but the equation themselves did not require
scaling to arrive at a correct result.
44

Example: Gauss Elimination
2 4
1 2 3 4
1 2 4
1 2 3 4
2 0
2 2 3 2 2
4 3 7
6 6 5 6
x x
x x x x
x x x
x x x x
a) Forward Elimination
0 2 0 1 0 6 1 6 5 6
2 2 3 2 2 2 2 3 2 2
1 4
4 3 0 1 7 4 3 0 1 7
6 1 6 5 6 0 2 0 1 0
R R
45

Example: Gauss Elimination (cont’d)
6 1 6 5 6
2 2 3 2 2 2 0.33333 1
4 3 0 1 7 3 0.66667 1
0 2 0 1 0
6 1 6 5 6
0 1.6667 5 3.6667 4
2 3
0 3.6667 4 4.3333 11
0 2 0 1 0
6 1 6 5 6
0 3.6667 4 4.3333 11
0 1.6667 5 3.6667 4
0 2 0 1 0
R R
R R
R R
46

6 1 6 5 6
0 3.6667 4 4.3333 11
0 1.6667 5 3.6667 4 3 0.45455 2
0 2 0 1 0 4 0.54545 2
6 1 6 5 6
0 3.6667 4 4.3333 11
0 0 6.8182 5.6364 9.0001
0 0 2.1818 3.3636 5.9999 4 0.32000 3
6 1 6
0 3.6667 4
0 0 6.8
0 0
R R
R R
R R
5 6
4.3333 11
182 5.6364 9.0001
0 1.5600 3.1199
47

b) Back Substitution
6 1 6 5 6
0 3.6667 4 4.3333 11
0 0 6.8182 5.6364 9.0001
0 0 0 1.5600 3.1199
4
3
2
1
3.1199
1.9999
1.5600
9.0001 5.6364 1.9999
0.33325
6.8182
11 4.3333 1.9999 4 0.33325
1.0000
3.6667
6 5 1.9999 6 0.33325 1 1.0000
0.50000
6
x
x
x
x 48

Gauss-Jordan Elimination
• It is a variation of Gauss elimination. The
major differences are:
– When an unknown is eliminated, it is eliminated
from all other equations rather than just the
subsequent ones.
– All rows are normalized by dividing them by their
pivot elements.
– Elimination step results in an identity matrix.
– It is not necessary to employ back substitution to
obtain solution.
49

Gauss-Jordan Elimination- Example
0 2 0 1 0 1 0.16667 1 0.83335 1
2 2 3 2 2 2 2 3 2 2
1 4
4 3 0 1 7 4 3 0 1 7
4/ 6.0
6 1 6 5 6 0 2 0 1 0
R R
R
1 0.16667 1 0.83335 1
2 2 3 2 2 2 2 1
4 3 0 1 7 3 4 1
0 2 0 1 0
R R
R R
50

1 0.16667 1 0.83335 1
0 1.6667 5 3.6667 2
0 3.6667 4 4.3334 7
0 2 0 1 0
Dividing the 2nd row by 1.6667 and reducing the second column.
(operating above the diagonal as well as below) gives:
1 0 1.5 1.2000 1.4000
0 1 2.9999 2.2000 2.4000
0 0 15.000 12.400 19.800
0 0 5.9998 3.4000 4.8000
Divide the 3rd row by 15.000 and make the elements in the 3rd
Column zero.
51
R1 – 0.16667 R2
R3 + 3.6667 R2
R4 – 2.0 R2
-4

1 0 0 0.04000 0.58000
0 1 0 0.27993 1.5599
0 0 1 0.82667 1.3200
0 0 0 1.5599 3.1197
Divide the 4th row by 1.5599 and create zero above the diagonal in the fourth
column.
1 0 0 0 0.49999
0 1 0 0 1.0001
0 0 1 0 0.33326
0 0 0 1 1.9999
Note: Gauss-Jordan method requires almost 50 % more operations than
Gauss elimination; therefore it is not recommended to use it.
52
R1 + 1.5 R3
R2 – 2.9999 R3
R4 + 5.9998 R3
R1 – 0.04000 R4
R2 + 0.27993 R4
R3 – 0.82667 R4

FG : 9.8, 9.9, 9.10
HW ; 9.6, 9.7, 9.9, 9.12, 9.13
53

Ch9-Gauss_Elimination4.pdf

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