1. Mujahed AlDhaifallah (Term 342) Read Chapter 9 of the textbook
EE 3561 : - Computational Methods Unit 3: Solution of Systems of Linear Equations
Mujahed AlDhaifallah
(Term 342)
Read Chapter 9 of the textbook
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2. Systems of linear equations
Coefficient
Matrix
Unknown
Vector
RHS
Vector
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3. Solutions of linear equations
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4. Solutions of linear equations
A set of equations is inconsistent if there exist no solution to the system of equations
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5. Solutions of linear equations
Some systems of equations may have infinite number of solutions
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6. Graphical Solution of Systems of Linear Equations
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7. Cramer’s Rule is not practical
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8. Lecture 6 Naive Gaussian Elimination
Examples
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9. Naive Gaussian Elimination
The method consists of two steps
Forward Elimination: the system is reduced to upper triangular form. A sequence of elementary operations is used.
Backward substitution: Solve the system starting from the last variable.
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10. Elementary Row operations
Adding a multiple of one row to another
Multiply any row by a non-zero constant
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11. Example Forward Elimination
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12. Example Forward Elimination
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13. Example Forward Elimination
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14. Example Backward substitution
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15. Forward Elimination
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16. Forward Elimination
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17. Backward substitution
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18. How many solutions does a system of equations AX=B have?
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19. How do we know if a solution is good or not
Given AX=B
X is a solution if AX-B=0
Due to computation error AX-B may not be zero
Compute the residuals R=|AX-B|
One possible test is ?????
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20. Determinant
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21. Linear Systems Types Singular Systems EE3561_Unit 3
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22. Detecting Singularity
After completing the elimination step, the determinant can be evaluated as the product of the diagonal elements.
One can detect singularity by the fact that the determinant of a singular system is zero.
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23. Problems with Naive Gaussian Elimination
The Naive Gaussian Elimination may fail for very simple cases. (The pivoting element is zero).
Very small pivoting element may result in , serious computation errors
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24. Possible solution
Equations Permutation: Naive Gaussian elimination method works well in the above two examples if the equations are first permuted, i.e., arranged as
The procedure that will do so is called “Gaussian elimination with scaled partial pivoting”.
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25. Gaussian Elimination with Scaled Partial Pivoting
In Gaussian elimination method, the order in which the equations are used as pivoting equations is the natural order {1, 2, 3, · · · , n}.
To overcome the problems that Naive Guassian elimination procedure face, we choose an order which is different than the natural used in forward elimination method. This is called partial pivoting.
The most important step in this method is to determine the pivot equation
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26. Gaussian Elimination with Scaled Partial Pivoting
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27. Scaled Partial Pivoting Procedure
Let l0 = {1, 2, 3, · · · , n}. This vector is called the “index vector”.
Define This vector is called the “Scale Vector” and is fixed for all operations. This means that si =absolute value of the maximum element in row i.
For iteration 1, define ratio#1 as
That is divide the absolute value of the elements of the first column by the corresponding elements in the “Scaled Vector”.
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28. Scaled Partial Pivoting Procedure
Choose the equation which give the greatest ratio as the first pivoting equation. Assume the greatest ratio is
Set the new index vector to be
i.e., interchange the place of 1 and i in l0 to get l1.
Then do the elimination as in the Naive Gaussian elimination method by taking raw i as the pivot raw and aii as the pivot element.
Repeat steps 3 to 6 for n − 1 iterations.
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29. Example 2
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30. Example 2 Initialization step
Scale vector:
disregard sign
find largest in magnitude in each row
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31. Why index vector?
Index vectors are used because it is much easier to exchange a single index element compared to exchanging the values of a complete row.
In practical problems with very large N, exchanging the contents of rows may not be practical since they could be stored at different locations.
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32. Example 2 Forward Elimination-- Step 1: eliminate x1
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33. Example 2 Forward Elimination-- Step 1: eliminate x1
First pivot equation
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34. Example 2 Forward Elimination-- Step 2: eliminate x2
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35. Example 2 Forward Elimination-- Step 2: eliminate x2
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36. Example 2 Forward Elimination-- Step 3: eliminate x3
Third pivot equation
Determinant =
(-1)3*(-1.5)(-2.5)(2)(4)
= 30-
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37. Example 2 Backward substitution
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38. Example 3
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39. Example 3 Initialization step
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40. Example 3 Forward Elimination-- Step 1: eliminate x1
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41. Example 3 Forward Elimination-- Step 1: eliminate x1
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42. Example 3 Forward Elimination-- Step 2: eliminate x2
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43. Example 3 Forward Elimination-- Step 2: eliminate x2
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44. Example 3 Forward Elimination-- Step 3: eliminate x3
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45. Example 3 Forward Elimination-- Step 3: eliminate x3
Determinant =
(-1)2*(0.8448)* ( )*(-10.5)*(4) =
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46. Example 3 Backward substitution
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47. How good is the solution?
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48. Remarks:
We use index vector to avoid the need to move the rows which may not be practical for large problems.
If you order equation as in the last value of the index vector, you have triangular form.
Scale vector is formed by taking maximum in magnitude in each row.
Scale vector does not change.
The original matrices A and B are used in Checking the residuals.
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