COMPUTATIONAL
MODELING FOR
ENGINEERING
MECN 6040
Professor: Dr. Omar E. Meza Castillo
omeza@bayamon.inter.edu
http://facultad.bayamon.inter.edu/omeza
Department of Mechanical Engineering
IMPLEMENTATION OF
SOLUTION METHODS
Gaussian elimination, LU
factorization, and others
INTRODUCTION
3
▪ 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.
SOLVING SMALL NUMBER 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.
4
▪ Nowadays, easy access to computers
makes the solution of large sets of linear
algebraic equations possible
and
practical.
SOLVING SMALL NUMBER
OF EQUATIONS
▪ There are many ways to solve a system of
linear equations:
▪ Graphical method
▪ Cramer’s rule
▪ Method of elimination
5
▪ Computer methods.
THE GRAPHICAL METHOD
 For two equations:
a11 x1  a12 x2  b1
a21 x1  a22 x2  b2
 Solve both equations for x2:
 a21 
b2
 x1 
x2  
a22
 a22 

x2  (slope) x1  intercept
6
 a11 
b1
 x1 
x2  
a12
 a12 
THE GRAPHICAL METHOD
 Or equate and solve for x1
 a11 
 a21 
b1
b2
 x1 
 x1 
x2  
 
a12
a22
 a12 
 a22 
 a21 a11 
b1 b2




x1 

0

a12 a22
 a22 a12 
7
 b1 b2   b2
b1 

 



a12 a22   a22 a12 

 x1  

 a21 a11   a21 a11 

 



 a22 a12   a22 a12 
THE GRAPHICAL METHOD
Infinite solutions
Ill-conditioned
(Slopes are too close)
8
No solution
9
THE GRAPHICAL METHOD
 2 x1  x 2  3
 x 2  2 x1  3
rearrange

x  x  3
x  3  x
2
1
 1
 2
x1 + x2 = 3
2x1 – x2 = 3
10
One solution
THE GRAPHICAL METHOD
No solution
2x1 – x2 = – 1
11
2x1 – x2 = 3
THE GRAPHICAL METHOD
Infinite solutions
6x1 – 3x2 = 9
12
2x1 – x2 = 3
CRAMER’S RULE
▪ Compute the determinant D
D
▪ 3 x 3 matrix
a11
a11
a21 a22
a12
a13
D  a21 a22
a31 a32
a23
a33
 a11
a12
a22
a23
a32
a33
 a12
 a11a22  a12a21
a21 a23
a31
a33
 a13
a21 a22
a31
a32
13
▪ 2 x 2 matrix
CRAMER’S RULE
▪ To find xk for the following system
a11 x1  a12 x2  ...  a1n xn  b1
a21 x1  a22 x2  ...  a2 n xn  b2

an1 x1  an 2 x2  ...  ann xn  bn
▪ Replace kth column of as with bs (i.e., aik  bi )
14
D( new matrix)
xk 
D(aij )
b1
D1 1
x1 
 b2
D D
b3
▪ 3 x 3 matrix
a11
a12
a13
D  a21 a22
a31 a32
a23
a33
a12
a22
a32
a13
a23
a33
a11 b1
D2 1
x2 
 a21 b2
D D
a31 b3
a13
a23
a33
a11 a12
D3 1
x3 
 a21 a22
D D
a31 a32
b1
b2
b3
15
CRAMER’S RULE
16
EXAMPLE 9.3
0.5 x1  x2  1.9 x3  0.67
0.1x1  0.3x2  0.5 x3  0.44
0.3 0.52
D  0.5
0.1
1
0.3
1
1.9
0.5
0.3  0.01 1
D2 1
x2 
 0.5 0.67 1.9  29.5
D D
0.1  0.44 0.5
0.3 0.52  0.01
D3 1
x3   0.5 1
0.67  19.8
D D
0.1 0.3  0.44
17
0.3x1  0.52 x2  x3  0.01
 0.01 0.52 1
D1 1
x1  
0.67
1 1.9  14.9
D D
 0.44 0.3 0.5
ELIMINATION METHOD
a11x1  a12 x2  b1

a21x1  a22 x2  b2
Eliminate x2 
Subtract to get
a22a11x1  a22a12 x2  a22b1

a12a21x1  a12a22 x2  a12b2
a22a11 x1  a12a21 x1  a22b1  a12b2
Not very practical for large number (> 4) of equations
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a22b1  a12b2
a11b2  a21b1
x1 
 x2 
a22a11  a12a21
a11a22  a12a21
▪ 2 x 2 matrix
 x1  2 x2  2
19
3x1  2 x2  18
2(18)  2(2)
x1 
4
3(2)  2(1)
3(2)  (1)(18)
x2 
3
3(2)  2(1)
NAVIE GAUSS ELIMINATION
The system can be written in a matrix format as
 Gauss elimination is the most
important algorithm to solve
systems of linear equations.
 It involves
equations
unknowns.
by
to
combining
eliminate
 a11 a12  a1n   x1   b1 
a a  a   x  b 
2n   2   2 
 21 22


     

   
an1 an 2  ann   xn  bn 
It involves 2 phases:
2. Back substitution: work from the last equation up.
20
1. Forward elimination phase: reduce the set of equations
to an upper triangular system.
NAVIE GAUSS ELIMINATION
 In the first step of the forward
elimination phase, x1 is eliminated from
all equations except the first one.
 The coefficient of x1 in the first
equation is called the pivot element.
 The second step is to eliminate x2 from
the third equation through the nth
equation.
 Do the same for all variables x3 to xn-1.
 The goal is to set up upper triangular
matrix
21
 The back-substitution phase starts from
the last equation up, to find the values
of x1, x2, …, xn.
NAVIE GAUSS ELIMINATION
Forward elimination phase
Gauss Elimination
Pseudocode
22
Back substitution
NAVIE GAUSS ELIMINATION
Use Gauss elimination to solve
Carry 6 significant figures
Solution
7.00333
x2  0.293333 x3  19.5617
23
1. Forward elimination: eliminate x1 from
equation (2):
 - (0.1/3)   
 0.1x1  0.00333333 x2  0.00666667 x3  0.261667
0.1x1  7
x2  0.3
x3  19.3
Eliminate
x1
from
 - (0.3/3)   
equation
(3):
After eliminating x1 from equations (2) and (3), the
system becomes
Eliminate x2 from equation (3):
 + (0.190000/7.00333)   
24
After eliminating x2 from equation (3), the system
becomes
2. Back Substitution: find the value of x3 from
equation (3):
70.0843
x3 
 7.0000
10.0120
Substitute the value of x3 in equation (2) to find the
value of x2:
7.00333x2  0.293333(7.0000)  19.5617
x2 
 19.5617  0.293333(7.0000)
 2.50000
7.00333
Substitute the values of x2 and x3 in equation (1) to
find the value of x1:
7.85  0.1(2.50000)  0.2(7.0000)
x1 
 3.00000
3
25
3x1  0.1(2.50000)  0.2(7.0000)  7.85
PITFALLS OF ELIMINATION
METHODS
 Division by zero (Partially solved by the
pivoting technique)
 Round-off errors (Important when large
number of equation are to be solved)
 Ill-conditioned systems:
Small changes in coefficients result in
large changes in the solution.
D  detA  0
26
When
TECHNIQUES FOR
IMPROVING SOLUTIONS
 Use of more significant figures (The
simplest remedy)
 Pivoting
(Determine
the
largest
available coefficient in the column
below the pivot element and switch
rows so that the largest element is the
pivot element)
0.0003x1  3.0000 x2  2.0001
27
1.0000 x1  1.0000 x2  1.0000
TECHNIQUES FOR
IMPROVING SOLUTIONS
▪ Scaling (Divide each row by the largest element
in that row)
x1 
With Scaling:
With Scaling and Pivoting:
x2  2
0.00002 x1  x2  1
x1  x2  2
x1  x2  2
0.00002 x1  x2  1
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Without Scaling:
2 x1  100,000 x2  100,000
29
MATLAB SCRIPT FILE: NGAUSSELIM.M
30
Example 9.5
>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]
A=
3.0000 -0.1000 -0.2000
0.1000 7.0000 -0.3000
0.3000 -0.2000 10.0000
>>Enter Solution Vector B > [7.85 -19.3 71.4]
B=
7.8500 -19.3000 71.4000
31
S=
3.0000
-2.5000
7.0000
32
MATLAB SCRIPT FILE: GAUSSELIMPPIVOT.M
Example
1x1  0 x2  2 x3  3 x4  1
 1x1  2 x2  2 x3  3 x4  1
0 x1  1x2  1x3  4 x4  2
6 x1  2 x2  2 x3  4 x4  1
>>Enter Matrix A > [1 0 2 3; -1 2 2 -3; 0 1 1 4; 6 2 2 4]
A=
1 0 2 3
-1 2 2 -3
0 1 1 4
6 2 2 4
33
>>Enter Solution Vector B > [1 -1 2 1]
B=
1 -1 2 1
S=
-0.1857
0.2286
-0.1143
0.4714
GAUSS-JORDAN METHOD
 It is a variation of Gauss elimination.
Both methods use row operations to
eliminate variables
 The major difference is that what an
unknown is eliminated, it is
eliminated from all other equations.
 Also, all rows (equations) are
normalized by dividing by their pivot
elements.
 The elimination phase produces an
identity matrix.
the
back
34
 It does not involve
substitution phase.
35
MATLAB SCRIPT FILE: GAUSSJORDAN.M
36
Example 9.12
>>Dame la matriz aumentada
Cuantas filas tiene la matriz: 3
Cuantas columnas tiene la matriz: 4
3.0 - 0.1
fila : 1
0.1 7.0
columna : 1Numero de esta fila y columna: 3

fila : 2
columna : 1Numero de esta fila y columna: 0.1
0.3 - 0.2
fila : 3
columna : 1Numero de esta fila y columna: 0.3 … Continuar
ingresando los valores de la matriz aumentada
a=
3.0000 -0.1000 -0.2000 7.8500
0.1000 7.0000 -0.3000 -19.3000
0.3000 -0.2000 10.0000 71.4000
a=
1.0000
0
0 3.0000
0 1.0000
0 -2.5000
0
0 1.0000 7.0000
- 0.2 7.85
- 0.3 - 19.3 
10.0 71.4
37
resultado
MATLAB’s Methods
To solve the system
Enter the following commands
>> A=[3 -.1 -.2; .1 7 -0.3;.3 -.2 10];
>> b=[7.85;-19.3;71.4];
>> inv(A)*b
ans =
38
3.0000
-2.5000
7.0000