Lecture 8 Gaussian Elimination

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Lecture 8 - Gaussian Elimination

• CVEN 302
• September 12, 2001

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Lectures Goals

• Discuss how to solve systems
• Gaussian Elimination
• Gaussian Elimination with Pivoting
• Tridiagonal Solver
• Problems with the technique
• Examples

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Gaussian Elimination

• There are two phases to the solving technique
• Elimination --- use row operations to convert the
  matrix into an upper diagonal matrix. (The
  elimination phase, which takes the the most
  effort and most susceptible to corruption by
  round off)
• Back substitution -- Solve x using a back
  substitution.

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Gaussian Elimination Algorithm

• Ax b
• Augment the n x n coefficient matrix with the
  vector of right hand sides to form a n x (n1)
• Interchange rows if necessary to make the value
  a11 with the largest magnitude of any coefficient
  in the first row
• Create zero in 2nd through nth row in first row
  by subtracting ai1 / a11 times first row from ith
  row

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Gaussian Elimination Algorithm

• Repeat (2) (3) for second through the (n-1)th
  rows, putting the largest magnitude coefficient
  in the diagonal by interchanging rows (consider
  only row j to n ) and then subtract times the
  jth row from the ith row so as to create zeros in
  all positions of jth column below the diagonal at
  conclusion of this step the system is upper
  triangular
• Solve for n from nth equation xn an,n1 / ann
• Solve for xn-1 , xn-2 , ...x1 from the (n-1)th
  through the first xi (ai,n1 - Sji1,n aj1
  xj ) / aii

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Example 1

• X1 3X2 5
• 2X1 4X2 6

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Example 2

• -3X1 2X2 - X3 -1
• 6X1 - 6X2 7X3 -7
• 3X1 - 4X2 4X3 -6

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Computer Program

• The program GEdemo(A,b) does the Gaussian
• elimination for a square matrix (n x n). It
• does not do any pivoting and works for only
• one b vector.

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Test the Program

• Example 1
• Example 2
• New Matrix
• 2X1 4X2 - 2 X3 - 2 X4 - 4
• 1X1 2X2 4X3 - 3 X4 5
• - 3X1 - 3X2 8X3 - 2X4 7
• - X1 X2 6X3 - 3X4
  7

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Problem with Gaussian Elimination

• The problem can occur when a zero appears in the
  diagonal and makes a simple Gaussian elimination
  impossible.
• Pivoting changes the matrix so that it will
  become diagonally dominate and reduce the
  round-off and truncation errors in the solving
  the matrix.

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Example of Pivoting

• 2 X1 4 X2 - 2 X3 10
• X1 2 X2 4 X3 6
• 2 X1 2 X2 1X3 2
• Answer X1 X2 X3 -3.40, 4.30, 0.20

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Computer Program

• GEPivotdemo(A,b) is a program, which will do a
  Gaussian elimination on matrix A with pivoting
  technique to make matrix diagonally dominate.
• The program is modification to handle a single
  value of b

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Question?

• How would you modify the programs to handle
  multiple inputs?
• What is diagonal matrix, upper triangular matrix,
  and lower triangular matrix?
• Can you do a column exchange and how would you
  handle the problem if it works?

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Question?

• What happens with the following example?
• 0.0001X1 0.5 X2 0.5
• 0.4000X1 - 0.3 X2 0.1
• What happens is the second equation becomes
  0.4000X1 - 2000 X2 -2000

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Gaussian Elimination

• If the diagonal is not dominate the problem can
  have round off error and truncation errors.
• The scaling will result in problems

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Question?

• What happens with the following example for
  values with two-significant figures?
• 0.4000 X1 - 0.3 X2 0.1
• 0.0001 X1 0.5 X2 0.5

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Scaling

• Scaling is an operation of adjusting the
  coefficients of a set of equations so that they
  are all of the same magnitude.
• A set of equations may involve relationships
  between quantities measured in a widely different
  units (N vs. kN, sec vs hrs, etc.) This may
  result in equation having very large number and
  others with very small , if we select pivoting
  may put numbers on the diagonal that are not
  large in comparison to other rows and create
  round-off errors that pivoting was suppose to
  avoid.

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Scaling

• What happens with the following example?
• 3X1 2 X2 100X3 105
• - X1 3 X2 100X3 102
• X1 2 X2 - 1X3 2

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Scaling

• The best way to handle the problem is to
  normalize the results.
• 0.03X1 0.02 X2 1.00X3 1.05
• - 0.01X1 0.03 X2 1.00X3 1.02
• 0.50X1 1.00 X2 - 0.50X3 1.00

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Gauss-Jordan Method

• The Gauss-Jordan Method is similar to the
  Gaussian Elimination.
• The method requires almost 50 more operations.

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Gauss-Jordan Method
The Gauss-Jordan method changes the matrix into
the identity matrix.
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Gauss-Jordan Method

• There are one phases to the solving technique
• Elimination --- use row operations to convert the
  matrix into an identity matrix.
• The new b vector is the solution to the x values.
  

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Gauss-Jordan Algorithm

• Ax b
• Augment the n x n coefficient matrix with the
  vector of right hand sides to form a n x (n1)
• Interchange rows if necessary to make the value
  a11 with the largest magnitude of any coefficient
  in the first row
• Create zero in 2nd through nth row in first row
  by subtracting ai1 / a11 times first row from ith
  row

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Gauss-Jordan Elimination Algorithm

• Repeat (2) (3) for first through the nth rows,
  putting the largest magnitude coefficient in the
  diagonal by interchanging rows (consider only row
  j to n ) and then subtract times the jth row
  from the ith row so as to create zeros in all
  positions of jth column and the diagonal becomes
  all ones
• Solve for all of the equations, xi ai,n1

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Example 1

• X1 3X2 5
• 2X1 4X2 6

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Example 2

• -3X1 2X2 - X3 -1
• 6X1 - 6X2 7X3 -7
• 3X1 - 4X2 4X3 -6

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Band Solver

• Large matrices tend to be banded, which means
  that the matrix has a band of non-zero
  coefficients and zeroes on the outside of the
  matrix.
• The simplest of the methods is the Thomas Method,
  which is used for a tridiagonal matrix.

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Advantages of Band Solvers

• The method reduce the number of operations and
  save the matrix in smaller amount of memory.
• The band solver is faster and is useful for large
  scale matrices.

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Thomas Method

• The method takes advantage of the bandedness of
  the matrix.
• The technique uses a two phase process.
• The first phase is to obtain the coefficients
  from the sweep.
• The second phase solves for the x values.

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Thomas Method

• The first phase starts with the first row of
  coefficients scales the a and r coefficients.
• The second phase solves for x values using the a
  and r coefficients.

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Thomas Method

• The program for the method is given as
  demoThomas(a,d,b,r)
• The algorithm is from the textbook, where a,d,b,
  r are vectors from the matrix.

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Summary

• Scaling of the problem will help in the
  convergence.
• Gauss-Jordan method is more computational intense
  and does not improve the round-off errors.
  However, it is useful for finding matrix
  inverses.
• Banded matrix solvers are faster and use less
  memory.

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Homework

• Check the Homework webpage