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The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Numerical Analysis
ECIV 3306
Chapter 9
Gauss Elimination
1
Associate Prof. Mazen Abualtayef Civil Engineering Department, The Islamic University of Gaza
Engineering Application for Part II Read 8.4 PIPE FRICTION
2
Bisection method
False position method
Newton-Raphson method
Fixed-point iteration method
Introduction
0)( xfRoots of a single equation:
A general set of equations: - n equations, - n unknowns.
0),,(
0),,(
0),,(
21
212
211
nn
n
n
xxxf
xxxf
xxxf
3
Linear Algebraic Equations
nnnnnn
nnx
nn
bxaxaxa
bxxaeaxa
bxaxxaxa
2211
23
22223
121
15
12112111
/)()(
)(
2
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
Nonlinear Equations
4
Review of Matrices
m n nm2n1n
m22221
m11211
aaa
aaa
aaa
]A[
2nd row
mth column
Elements are indicated by a i j
row column
Row vector: Column vector:
n 1 n21 rrr]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.
5
Review of Matrices
Special types of square 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 ij • Identity matrix: [A]nxn is an identity matrix if it is diagonal matrix with aii=1 i=1,...,n . Shown as [I]
6
Review of Matrices
• Upper triangular matrix:
[A]nxn is upper triangular if aij=0 i=1,...,n ; j=1,...,n and i>j
All the elements below the main diagonal are zero
• 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: تبديل
[B] is the transpose of [A]nxn if bij=aji Shown as [A] or [A]T
7
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
aa
aaa
A
][222
11211
nnn a a
aa
a
A
1
2221
11
][
Upper Triangular Lower Triangular
8
Review of Matrices
• Matrix multiplication:
r
1k
kjikij b a cNote: [A][B] [B][A]
9
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|.
2221
1211
aa
aaA
2
1
b
bB
22221
11211
baa
baa
10
Part 3- Objectives
11
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
12
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.
2222121
1212111
bxaxa
bxaxa
22
21
22
212
12
12
11
12
112 intercept(slope)
a
bx
a
ax
xxa
bx
a
ax
13
Graphical Method -Example
• Solve:
• Plot x2 vs. x1, the intersection of the lines presents the solution.
22
1823
21
21
xx
xx
14
Graphical Method
No solution Infinite solution ill condition (sensitive to round-off errors)
15
Graphical Method
Mathematically
• Coefficient matrices of (a) & (b) are singular. There is no unique solution for these systems. Determinants of the coefficient matrices are zero and these matrices can not be inverted.
• Coefficient matrix of (c) is almost singular. Its inverse is difficult to take. This system has a unique solution, which is not easy to determine numerically because of its extreme sensitivity 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:
BxA .
333231
232221
131211
aaa
aaa
aaa
A
17
Cramer’s Rule
22313221
3231
2221
13
23313321
3331
2321
12
23323322
3332
2322
11
333231
232221
131211
aaaaaa
aaD
aaaaaa
aaD
aaaaaa
aaD
aaa
aaa
aaa
D
3231
2221
13
3331
2321
12
3332
2322
11aa
aaa
aa
aaa
aa
aaaD
D
aab
aab
aab
x33323
23222
13121
1
D
aba
aba
aba
x33331
23221
13111
2
D
baa
baa
baa
x33231
22221
11211
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.
19
3. Method of Elimination
• The basic strategy is to successively solve one of the equations of the set for one of the unknowns and to eliminate that variable from the remaining equations by substitution.
• 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.
20
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).
21
9.2 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.
22
Naive Gauss Elimination Method • 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.
23
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
nn3n2n
n33332
n22322
n1131211
b
b
b
b
x
x
x
x
aaa0
aaa0
aaa0
aaaa
Naive Gauss Elimination Method (cont’d)
24
pivot محور
Naive Gauss Elimination Method (cont’d)
1.2- Now eliminate x2 from 3rd to nth equations.
The modified system is
n
3
2
1
n
3
2
1
nn3n
n333
n22322
n1131211
b
b
b
b
x
x
x
x
aa00
aa00
aaa0
aaaa
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
25
Naive Gauss Elimination Method (cont’d)
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
nn
a
bx
121for )1(
1
1)1(
, ..., , n-n- ia
xab
xi
ii
n
ij
j
i
ij
i
i
i
26
Summary of Naive Gauss Elimination Method
27
Pseudo-code of Naive Gauss Elimination Method
(a) Forward Elimination (b) Back substitution
k,j
28
Naive Gauss Elimination Method
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
Naive Gauss Elimination Method
Example 1 (cont’d)
Step 1: Forward elimination 1. Eliminate x1 6 –2 2 4 | 16 (Does not change. Pivot is 6) 0 –4 2 2 | -6
0 –12 8 1 | -27 R3-3R2 0 2 3 -14 | -18 R4-(-0.5R2)
2. Eliminate x2 6 –2 2 4 | 16 (Does not change.) 0 –4 2 2 | -6 (Does not change. Pivot is-4) 0 0 2 -5 | -9
0 0 4 -13 | -21 R4-2R3
3. Eliminate x3 6 –2 2 4 | 16 (Does not change.) 0 –4 2 2 | -6 (Does not change.) 0 0 2 -5 | -9 (Does not change. Pivot is 2) 0 0 0 -3 | -3
30
Naive Gauss Elimination Method 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 .
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 + 3.000 x2 = 2.0001
1.0000 x1 + 1.000 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
22 x
0003.0
)3/2(30001.21
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 previous 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 sig x2 x1 Ea% in x1
4 0.6667 0.3333 0.01
5 0.66667 0.33333 0.001
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
Determinate Evaluation
The determinate can be simply evaluated at the end of the forward elimination step, when the program employs partial pivoting:
45
' '' ( 1)
11 22 33
where:
represents the number of times that ro
1
ws are pivoted
pn
nnD a a a a
p
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 21 4
4 3 0 1 7 4 3 0 1 7
6 1 6 5 6 0 2 0 1 0
R R
46
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 42 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
Example: Gauss Elimination (cont’d)
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
48
Example: Gauss Elimination (cont’d)
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
3.11991.9999
1.5600
9.0001 5.6364 1.99990.33325
6.8182
11 4.3333 1.9999 4 0.333251.0000
3.6667
6 5 1.9999 6 0.33325 1 1.00000.50000
6
x
x
x
x
49
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.
50
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 21 4
4 3 0 1 7 4 3 0 1 74 / 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
51
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.
52
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.
53