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Gaussian elimination

  1. Jan 19, 2005 #1
    I am not sure how to solve this:
    Given an augmented matrix, find conditions on a, b, c for which the system has solutions:
    Code (Text):

    -1   -2   3    b
    -1   -6   23   c
    -3   2    4    a
     
    so by Gaussian elimination, the matrix I ended up with is
    Code (Text):

    1   2   -3   -b
    0   4   -20  b-c
    0   0   -35  (3b-a) + 2(b-c)
     
    And now I am stuck, because there is still a variable left in the last row. Does that mean there are no such a, b, c for which the system has a solution?
    Thanks in advance.
     
  2. jcsd
  3. Jan 19, 2005 #2

    learningphysics

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    Homework Helper

    In general, when do you know there is a solution, and when do you know there is no solution?

    Suppose you were given a particular a,b and c at random... would you be able to solve for x, y and z?
     
  4. Jan 19, 2005 #3
    So, are you implying that the conditions I need on a, b, and c are such that
    z = (3b - a + 2b - 2c)/ (-35) and so on?
    The problems we have done in class were basically of the form such that the last row of coefficient part of the augm. matrix is all zeros without variables. So the conditions are only in terms of a, b, and c themselves.
     
  5. Jan 19, 2005 #4

    learningphysics

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    The idea is that no matter what a,b,c are you can always get a solution. Like you showed above you can always solve for z, then for y then for x.

    a, b and c can be any real number.

    Another way to look at it is, after you reduce an augmented matrix to echelon form.. the only time you have no solution is if you have a row that is

    0 0 0 | t

    where t is non-zero. Any other time there is always a solution.

    Since your matrix (reduced to echelon form) does not have a row like this (no matter what values a, b or c take) there is a solution for any a,b,c.

    So a,b,c can be any real numbers.
     
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