Numerical Analysis

1. Numerical Methods
Dr. Md. Abdul Hasib
Associate Professor
Department of Mechanical Engineering
Khulna University of Engineering & Technology (KUET)

2. Content
References
You may follow any book, whatever you want
 Introductory Methods of Numerical Analysis
S. S. Sastry
 Numerical Methods
E-Balaguruswamy
 Numerical Methods for Engineers
S. C. Chapra and R. P. Canale
o Internet

3. Content
For the Students
3
 Always come in time
 Be attentive
 Ask questions
 If you face any problem, never hesitate to contact with me
You may contact through…
Email or Mobile Phone
ahasib@me.kuet.ac.bd

4. CSE-2303
Numerical Methods
Solution of Linear Algebraic Equations

5. 5
Solution of Linear Algebraic Equations
Solution of linear algebraic equations
 Indirect Methods
 Jacobi Method
 Gauss-Seidel Method
 Conjugate Gradient Method
 Direct Method
 Gauss Elimination Method
 Gauss Jordan Method
 Matrix Inversion Method
 LU Decomposition Method
Solve:

6. 6
n 0 1 2 3 4 5 6 7
0.0 -0.200 0.146 0.192 0.181 0.185 0.186 0.186
0.0 0.222 0.203 0.328 0.332 0.329 0.331 0.331
0.0 -0.429 -0.517 -0.416 0.421 -0.424 -0.423 -0.423
Jacobi Method

7. 7
n 0 1 2 3 4 5
0.0 -0.200 0.167 0.191 0.186 0.186
0.0 0.156 0.334 0.333 0.331 0.331
0.0 -0.508 -0.429 -0.429 -0.423 -0.423
Gauss-Seidel Method

8. 8
n 0 1 2 3 4 5
0.0 -4 -34 -194 -6124 -42874
0.0 -6 -34 -244 -8574 -42874
Limitations of Jacobi & Gauss-Seidel Method
n 0 1 2 3 4 5
0.0 -4 -174 -6124 -214374 -7503124
0.0 -34 -1224 --42874 -1580624 -52521874

9. 9
n 0 1 2 3 4 5
0.0 0.8571 0.9959 0.9999 1.0000 1.0000
0.0 0.9714 0.9991 1.0000 1.0000 1.0000
Pivoting
• Partial Pivoting
• Complete Pivoting
Limitations: How to Solve

10. Gauss Elimination Method
A method to solve simultaneous linear equations of the form
[A][X]=[C]
Two steps:
1. Forward Elimination
2. Back Substitution

11. A set of n equations and n unknowns
1
1
3
13
2
12
1
11 ... b
x
a
x
a
x
a
x
a n
n 




2
2
3
23
2
22
1
21 ... b
x
a
x
a
x
a
x
a n
n 




n
n
nn
n
n
n b
x
a
x
a
x
a
x
a 



 ...
3
3
2
2
1
1
. .
. .
. .
(n-1) steps of forward elimination
Forward Elimination

12. Step 1
For Equation 2, divide Equation 1 by and
multiply by .
)
...
( 1
1
3
13
2
12
1
11
11
21
b
x
a
x
a
x
a
x
a
a
a
n
n 










1
11
21
1
11
21
2
12
11
21
1
21 ... b
a
a
x
a
a
a
x
a
a
a
x
a n
n 



11
a
21
a
Forward Elimination

13. Forward Elimination
1
11
21
1
11
21
2
12
11
21
1
21 ... b
a
a
x
a
a
a
x
a
a
a
x
a n
n 



2
2
3
23
2
22
1
21 ... b
x
a
x
a
x
a
x
a n
n 




1
11
21
2
1
11
21
2
2
12
11
21
22 ... b
a
a
b
x
a
a
a
a
x
a
a
a
a n
n
n 





















'
2
'
2
2
'
22 ... b
x
a
x
a n
n 


Subtract the result from Equation 2.
−
_________________________________________________
or

14. Repeat this procedure for the remaining
equations to reduce the set of equations as
1
1
3
13
2
12
1
11 ... b
x
a
x
a
x
a
x
a n
n 




'
2
'
2
3
'
23
2
'
22 ... b
x
a
x
a
x
a n
n 



'
3
'
3
3
'
33
2
'
32 ... b
x
a
x
a
x
a n
n 



'
'
3
'
3
2
'
2 ... n
n
nn
n
n b
x
a
x
a
x
a 



. . .
. . .
. . .
End of Step 1
Forward Elimination

15. Step 2
Repeat the same procedure for the 3rd term of
Equation 3.
1
1
3
13
2
12
1
11 ... b
x
a
x
a
x
a
x
a n
n 




'
2
'
2
3
'
23
2
'
22 ... b
x
a
x
a
x
a n
n 



"
3
"
3
3
"
33 ... b
x
a
x
a n
n 


"
"
3
"
3 ... n
n
nn
n b
x
a
x
a 


. .
. .
. .
End of Step 2
Forward Elimination

16. At the end of (n-1) Forward Elimination steps, the
system of equations will look like
'
2
'
2
3
'
23
2
'
22 ... b
x
a
x
a
x
a n
n 



"
3
"
3
3
"
33 ... b
x
a
x
a n
n 


   
1
1 


n
n
n
n
nn b
x
a
. .
. .
. .
1
1
3
13
2
12
1
11 ... b
x
a
x
a
x
a
x
a n
n 




End of Step (n-1)
Forward Elimination

17. Matrix Form at End of Forward Elimination

















































 )
(n-
n
"
'
n
)
(n
nn
"
n
"
'
n
'
'
n
b
b
b
b
x
x
x
x
a
a
a
a
a
a
a
a
a
a
1
3
2
1
3
2
1
1
3
33
2
23
22
1
13
12
11
0
0
0
0
0
0
0










Forward Elimination

18. Back Substitution
Solve each equation starting from the last equation
Example of a system of 3 equations

19. Back Substitution Starting Eqns.
'
2
'
2
3
'
23
2
'
22 ... b
x
a
x
a
x
a n
n 



"
3
"
3
"
33 ... b
x
a
x
a n
n 


   
1
1 


n
n
n
n
nn b
x
a
. .
. .
. .
1
1
3
13
2
12
1
11 ... b
x
a
x
a
x
a
x
a n
n 





20. Back Substitution
Start with the last equation because it has only one unknown
)
1
(
)
1
(


 n
nn
n
n
n
a
b
x

21. Back Substitution
       
  1
,...,
1
for
...
1
1
,
2
1
2
,
1
1
1
,
1






 








n
i
a
x
a
x
a
x
a
b
x i
ii
n
i
n
i
i
i
i
i
i
i
i
i
i
i
i
   
  1
,...,
1
for
1
1
1
1




 




n
i
a
x
a
b
x i
ii
n
i
j
j
i
ij
i
i
i
)
1
(
)
1
(


 n
nn
n
n
n
a
b
x

22. Gauss Elimination Method-Example
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Back Substitution
z= 3, y=2 and x=1

23. Limitations: Gauss Elimination Method
0.w+2x – y + 3z = 9
2w+x + y + z = 6
3w-x – y + z = 2
5w+3x-5y+7z=10
Pivoting
• Partial Pivoting
2w+x + y + z = 6
0.w+2x – y + 3z = 9
3w-x – y + z = 2
5w+3x-5y+7z=10
2x +0.w– y + 3z = 9
x + 2w+ y + z = 6
-x +3w- y + z = 2
3x+5w+5y+7z=10
0.w+0.x – y + 3z = 9
0.w+x + y + z = 6
3w-x – y + z = 2
5w+3x-5y+7z=10
Pivoting
• Complete Pivoting
x + 0.w+ y + z = 6
0.x +0.w– y + 3z = 9
-x+3w – y + z = 2
3x+5w-5y+7z=10
OR

24. Gauss Jordan Method
A method to solve simultaneous linear equations of the form
[A][X]=[C]

25. Gauss Jordan Method-Example
z= 3, y=2 and x=1
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9

26. Gauss Jordan Method-Example

27. Gauss Jordan Method-Example

28. Matrix Inversion Method

29. Matrix Inversion Method

30. Matrix Inversion Method

31. Matrix Inversion Method

32. Matrix Inversion Method

33. Home Work - Solution of Linear Algebraic Equations
x - 3y + 12z = 31
4x + y - z = 3
2x + 7y + z = 19
Solve the above equations using
• Jacobi Method
• Gauss-Seidel Method
• Gauss Elimination Method
• Gauss Jordan Method
• Matrix Inversion Method

34. Thank you very much
সবাইকে অকেে ধেযবাদ