1. Chapter 9
Gauss Elimination

2. Gauss Elimination 9.1 Solving small numbers of equations
9.2 Naive Gauss Elimination
9.3 Pivoting
9.4 Tridiagonal Systems
MATLAB M-files
GaussNaive, GaussPivot, Tridiag

3. Small Matrices Graphical Cramer's rule Elimination
For small numbers of equations, can be solved by hand
Graphical
Cramer's rule
Elimination

4. Graphical Method
x1 + x2 = 3
2x1 – x2 = 3
One solution

5. Graphical Method
No solution
2x1 – x2 = – 1
2x1 – x2 = 3

6. Graphical Method
Infinite many solution
2x1 – x2 = 3
6x1 – 3x2 = 9

7. Graphical Method
Ill conditioned
2x1 – x2 = 3
2.1x1 – x2 = 3

8. Cramer’s Rule
Compute the determinant D
2 x 2 matrix
3 x 3 matrix

9. Cramer’s Rule To find xk for the following system
Replace kth column of as with bs (i.e., aik  bi )

10. Example
3 x 3 matrix

11. Ill-Conditioned System
What happen if the determinant D is very small or zero?
Divided by zero (linearly dependent system)
Divided by a small number: Round-off error
Loss of significant digits

12. Elimination Method Eliminate x2  Subtract to get
Not very practical for large number (> 4) of equations

13. MATLAB’s Methods Forward slash ( / ) Back-slash ( \ )
Multiplication by the inverse of the quantity under the slash

14. Gauss Elimination
Manipulate equations to eliminate one of the unknowns
Develop algorithm to do this repeatedly
The goal is to set up upper triangular matrix
Back substitution to find solution (root)

15. Basic Gauss Elimination
Direct Method (no iteration required)
Forward elimination
Column-by-column elimination of the below-diagonal elements
Reduce to upper triangular matrix
Back-substitution

16. Naive Gauss Elimination
Begin with
Multiply the first equation by a21 / a11 and subtract from second equation

17. Gauss Elimination
Reduce to
Repeat the forward elimination to get

18. Gauss Elimination First equation is pivot equation
a11 is pivot element
Now multiply second equation by a'32 /a'22 and subtract from third equation

19. Gauss Elimination Repeat the elimination of ai2 and get
Continue and get

20. Back Substitution Now we can perform back substitution to get {x}
By simple division
Substitute this into (n-1)th equation
Solve for xn-1
Repeat the process to solve for xn-2 , xn-3 , …. x2, x1

21. Back Substitution Back substitution: starting with xn
Solve for xn1 , xn2 , … , 3, 2, 1
for i = n1, n2, …, 1
Naive Gauss Elimination

22. Elimination of first column

23. Elimination of second column

24. Elimination of third column
Upper triangular matrix

25. Back-Substitution
Upper triangular matrix

26. Example

27. Forward Elimination

28. Upper Triangular Matrix

29. Back-Substitution

30. MATLAB Script File: GaussNaive

31. Print all factor and Aug (do not suppress output)
>> format short
>> x = GaussNaive(A,b)
m = 4
n =
Aug =
factor = -1 6
factor =
0.5000
Aug = 1 14
x =
0.4714
0.2286 x4
Aug = [A, b] x3 x2
Eliminate second column x1
Back-substitution
Eliminate third column
Print all factor and Aug
(do not suppress output)
Eliminate first column

32. Algorithm for Gauss elimination
1. Forward elimination
for each equation j, j = 1 to n-1
for all equations k greater than j
(a) multiply equation j by akj /ajj
(b) subtract the result from equation k
This leads to an upper triangular matrix
2. Back-Substitution
(a) determine xn from
(b) put xn into (n-1)th equation, solve for xn-1
(c) repeat from (b), moving back to n-2, n-3, etc until all equations are solved

33. Operation Count Important as matrix gets large For Gauss elimination
Elimination routine uses on the order of O(n3/3) operations
Back-substitution uses O(n2/2)

34. Operation Count
Total operation counts for elimination stage = 2n3/3 + O(n2)
Total operation counts for back substitution stage = n2 + O(n)

35. Operation Count
Number of flops (floating-point operations) for Naive Gauss elimination
Computation time increase rapidly with n
Most of the effort is incurred in the elimination step
Improve efficiency by reducing the elimination effort

36. Partial Pivoting Problems with Gauss elimination
division by zero
round off errors
ill conditioned systems
Use “Pivoting” to avoid this
Find the row with largest absolute coefficient below the pivot element
Switch rows (“partial pivoting”)
complete pivoting switch columns also (rarely used)

37. Round-off Errors A lot of chopping with more than n3/3 operations
More important - error is propagated
For large systems (more than 100 equations), round-off error can be very important (machine dependent)
Ill conditioned systems - small changes in coefficients lead to large changes in solution
Round-off errors are especially important for ill-conditioned systems

38. Ill-conditioned System
2x1 – x2 = 3
2.1x1 – x2 = 3

39. Ill-Conditioned System
Consider
Since slopes are almost equal
Divided by small number

40. Determinant
Calculate determinant using Gauss elimination

41. Gauss Elimination with Partial Pivoting
Forward elimination
for each equation j, j = 1 to n-1
first scale each equation k greater than j
then pivot (switch rows)
Now perform the elimination
(a) multiply equation j by akj /ajj
(b) subtract the result from equation

42. Partial (Row) Pivoting

43. Forward Elimination
Interchange rows 1 & 4

44. Forward Elimination
No interchange required

45. Back-Substitution
Interchange rows 3 & 4

46. MATLAB M-File: GaussPivot
Partial Pivoting (switch rows)
largest element in {x}
[big,i] = max(x)
Partial Pivoting
index of the largest element

47. Eliminate first column
>> format short
>> x=GaussPivot0(A,b)
Aug =
big = 6
i = 4
ipr =
factor =
0.1667
Aug = [A b]
Find the first pivot element and its index
Interchange rows 1 and 4
Eliminate first column
No need to interchange

48. Save factors fij for LU Decomposition
Back substitution
big =
2.3333
i = 1
ipr = 2
factor =
0.4286
Aug = 4
Second pivot element and index
No need to interchange
x =
0.4714
0.2286
Eliminate second column
Third pivot element and index
Interchange rows 3 and 4
Save factors fij for
LU Decomposition
Eliminate third column

49. TRUSS 4 5 F45 F35 F14 F24 F25   1 3 H1 2 F23 F12 V1 V3  
W = 100 kg

50. Statics: Force Balance
Node 1
Node 2
Node 3
Node 4
Node 5

51. Example: Forces in a Simple Truss

52. Define Matrices A and b in script file
function [A,b]=Truss(alpha,beta,gamma,delta)
A=zeros(10,10);
A(1,1)=1; A(1,5)=sin(alpha);
A(2,2)=1; A(2,4)=1; A(2,5)=cos(alpha);
A(3,7)=sin(beta); A(3,8)=sin(gamma);
A(4,4)=-1; A(4,6)=1; A(4,7)=-cos(beta); A(4,8)=cos(gamma);
A(5,3)=1; A(5,9)=sin(gamma);
A(6,6)=-1; A(6,9)=-cos(delta);
A(7,5)=-sin(alpha); A(7,7)=-sin(beta);
A(8,5)=-cos(alpha); A(8,7)=cos(beta); A(8,10)=1;
A(9,8)=-sin(gamma); A(9,9)=-sin(delta);
A(10,8)=-cos(gamma); A(10,9)=cos(delta); A(10,10)=-1;
b=zeros(10,1); b(3,1)=100;

53. Gauss Elimination with Partial Pivoting
>> alpha=pi/6; beta=pi/3; gamma=pi/4; delta=pi/3;
>> [A,b] = Truss(alpha,beta,gamma,delta)
A =
b =
100
>> x = GaussPivot(A,b)
x =
Simple truss

54. Banded Matrix
HBW: Half Band Width

55. Banded Matrix ai,j= 0 if j > i + HB or j< i - HB
HB: Half Bandwidth
B: Bandwidth
B = 2*HB + 1
In this example
HB = 1 & B = 3

56. Tridiagonal Matrix
Only three nonzero elements in each equation (3n instead of n2 elements)
Subdiagonal, diagonal, superdiagonal
Row Scaling (not implemented in textbook)
-- scale the diagonal element to aii = 1
Solve by Gauss elimination

57. Tridiagonal Matrix Special case of banded matrix with bandwidth = 3
Save storage, 3  n instead of n  n

58. Tridiagonal Matrix Forward elimination Back substitution
Use factor = ek / fk1
to eliminate subdiagonal element
Apply the same matrix operations to right hand side

59. Hand Calculations: Tridiagonal Matrix
(a) Forward elimination (b) Back substitution

60. MATLAB M-file: Tridiag

61. Example: Tridiagonal matrix
function [e,f,g,r] = example
e=[ ];
f=[ ];
g=[ ];
r=[ ];
» [e,f,g,r] = example
e =
f =
g =
r =
» x = Tridiag (e, f, g, r)
x =
Note: e(1) = 0 and g(n) = 0

62. Chapter 8 Problem 8.2 (20), 8.3(10), 8.7(15) Chapter 9
CVEN Homework No. 6
Chapter 8
Problem 8.2 (20), 8.3(10), 8.7(15)
Chapter 9
Problem 9.5 (15), 9.8(20), 9.11(20).
Due Monday 10/06/08 at the beginning of the period