Lecture 29 Outline

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• Finish Last Lecture

• Some difficulties that can arise with linear equations

  The normal situation with n equations in n unknowns is that we will find exactly one set of values for the variables that satisfy the equations. Unfortunately, this is not always the case. There are two possible problems:

  • There may be no solutions at all. (Inconsistent equations)

    Example:

    2x + 3y = 1
    2x + 3y = 2

  • There may be infinitely many solutions. (Underdetermined equations)

    Example:

    2x + 3y = 1
    4x + 6y = 2
    

    Here any solution of the first equation is automatically a solution of the second because the second is just double the first. To satisfy the first, take any value of y whatsoever, and then x = (1-3y)/2.

• More on Gaussian Elimination

  The basic step in Gaussian elimination can be described as follows:

  1. Select a nonzero entry in the matrix.
  2. Convert (using a row operation) the entry to a 1.
  3. Convert (" " ") all other entries in that column to zeros

  This is called a pivot operation and the selected entry is called the pivot. (There are geometric reasons for this terminology we can't go into here.)

  Example: Pivot around the upper left corner entry.

  2	-2	4	|	6
  3	-1	1	|	2
  -3	1	-1	|	-2

  1	-1	2	|	3
  3	-1	1	|	2
  -3	1	-1	|	-2

  1	-1	2	|	3
  0	2	-5	|	-7
  0	-2	5	|	7

  Our final version of the Gaussian elimination procedure now reads:

  1. Write down the augmented matrix
  2. Interchange rows if needed so the 1,1 entry is nonzero
  3. Pivot around the 1,1 entry
  4. Interchange rows if needed so the 2,2 entry is nonzero
  5. Pivot around the 2,2 entry
  6. Continue as long as possible, or until every diagonal entry has been converted to a 1.

• Solving underdetermined systems

  Let's continue the procedure with the example above. Pivoting around the entry 2 in the second row and second column, we obtain:

  1	-1	2	|	3
  0	1	-2.5	|	-3.5
  0	-2	5	|	7

  1	0	-.5	|	-.5
  0	1	-2.5	|	-3.5
  0	0	0	|	0

  The procedure cannot continue because there is no nonzero entry in the last row. The equations that correspond to this last augmented matrix are

  x - (1/2)z = -1/2
  y - (5/2)z = -7/2
  0 = 0

  There are infinitely many solutions, because z can be picked to have any value whatsoever, and then x and y expressed in terms of the chosen z. We could express this solution as:

  x = -1/2 + 1/2 z
  y = -7/2 + 5/2 z
  z arbitrary

  In general, if there were as many equations as unknowns, and if M zero rows appear at the bottom at the end of the procedure, then M of the variables can be chosen arbitrarily and the remaining variables expressed in terms of them.

• Inconsistent Systems

  Suppose we changed the final 7 in the original system to an 8. Then the last step would have ended with:

  1	0	-.5	|	-.5
  0	1	-2.5	|	-3.5
  0	0	0	|	1

  This time the last row corresponds to the impossible equation 0 = 1. When this happens it is a signal that the original system was inconsistent , i.e., there were no solutions.

• Row Echelon Form

  The final matrix in the Gaussian elimination procedure has a special format called row echelon form. We say a column is in proper form if has a single entry of 1 and all other entries are 0. It is always possible to achieve a row echelon form in which N-M columns are in proper form and the remaining M columns are not (M may be zero.) Also, the last M rows are all 0s.

• Good news about the TI-83

  The good news is that the calculator will convert an augmented matrix to row echelon form in 1 step! Use the rref command on the MATRIX/MATH menu.

  Example: Solve by reducing to row echelon form (2.2.19):

  x + y - 2z + 2w = 5
  2x + y - 4z + w = 5
  3x + 4y - 6z + 9w = 20
  4x + 4y - 8z + 8w = 20

• Additional problems.