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Constants in a system of equations that makes the system consistent

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    system.jpg

    3. The attempt at a solution

    For a) I just solved simultaneous equations, a link to a resource that solves a system more efficiently would be nice but augmented matrices are taught in the syllabus, or atleast not in the official text. Although I've seen solutions in the mark schemes of augmented matrices.

    Next I get the determinent in terms of a, a=1 when determinent is 0.

    I'm not sure how to do this last part.
     
  2. jcsd
  3. Jun 16, 2009 #2

    Pengwuino

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    Gold Member

    If you conduct row operations, you would have gotten the same result for part b) but in the form of a row being something like (a-*)x + 0y + 0z = b+** with * and ** being whatever numbers you actually get, I didn't calculate it. Well what you found was that * was equal to 1 to make it 0x + 0y + 0z = something which was inconsistent and had no solution. However, that "something" which is a function of b can be made to be 0 depending on what you pick for b. That is, you'd get 0x + 0y + 0z = 0 which is now a consistent system
     
  4. Jun 16, 2009 #3

    EnumaElish

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    When a = 1, does the system have any solution?

    Is that what they meant by "does not have a unique solution"?
     
  5. Jun 16, 2009 #4
    No solutions or infinite solutions i.e. a line of solutions in the intersection of two planes.
     
  6. Jun 16, 2009 #5

    Pengwuino

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    I assume the spirit of the problem is to first find a "no solution" then an infinite solution in part ii) since in part i), you'll have b that isn't known yet.
     
  7. Jun 16, 2009 #6
    I'm trying to work out how to do what Pengwuino said for me to do.
     
  8. Jun 16, 2009 #7

    Mark44

    Staff: Mentor

    To summarize part of what Pengwuino said, after row reducing the augmented matrix [A|b], where A represents your matrix of coefficients, and b represents the vector of constants on the right sides of the equations, you're looking to get one or more rows of zeroes in the bottom of A.

    There are two possibilities:

    1. 0 0 0 | k , where k != 0
      This represents 0x + 0y + 0z = k, for which there is no solution.
    2. 0 0 0 | 0
      This represents 0x + 0y + 0z = 0, for which there are an infinite number of solutions.
     
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