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Homework Help: Linear Algebra -

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Existence and Uniqueness of Solutions and Pivot Structure. Determine the values of h and k so that,
    [1 3 | 2]
    [3 h | k]

    (the above is meant to be 1 augmented matrix)

    (1) Is consistent (has a solution).
    (2) Is consistent with a unique solution.
    (3) Is inconsistent.

    2. Relevant equations
    Row reduction?

    3. The attempt at a solution

    I have no idea where to even begin, I haven't learned anything during class and I can't find any help in the book. If someone could help me get started or preferably tell me the steps of how i would go about doing this I can try it out and post my work here. Thank you
  2. jcsd
  3. Jul 3, 2010 #2


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    I would recommend using row reduction to solve the system. Of course your solution will depend on h and k.

    What values of h and k is it impossible to have a solution? (Probably because a denominator is 0 but the numerator is not.)

    What values of h and k give an infinite number of solutions? (Probably because both numerator and denominator of a fraction are 0.)

    All other values of h and k will give a unique solution.

    (If it is really true that you "haven't learned anything during class" and "can't find any help in the book", you are probably in a class you are not ready for and should consider dropping the class.)
  4. Jul 3, 2010 #3
    thanks for the help, i'll try that. Its not that i'm not ready for the class, the teacher doesn't give many examples at all its just theorem after theorem with him and no explination. Everyone I know in the class is having the same issue. And the book doesn't have examples similar to things he gives on the homework, I mean they're somewhat similar but not enough to learn off of and do it correctly
  5. Jul 3, 2010 #4
    here's what I did so far.. I did r2 = r2 - 3r1 and got
    [1 3 | 2 ]
    [0 h-9 | k-6 ]
    so I have
    x1+3x2 = 2 and
    (h-9)x2 = k-6 so...
    x2 = (k-6)/(h-9)

    Hopefully its right upto this point

    So my inconsistent solution would be when h = 9
    I'm still a little confused about the consistent and unique
  6. Jul 3, 2010 #5


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    Ack; you can't divide by an unknown number! What if it happened to be zero?!?!
  7. Jul 3, 2010 #6
    I think thats the point, Its not consistent if that unknown number happens to be what it needs to be zero in the denominator
  8. Jul 3, 2010 #7


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    You don't know that. You are drawing your conclusion from the results of division -- but you weren't allowed to divide in the first place, so you don't know if your conclusion is right!
  9. Jul 3, 2010 #8
    hmm well my teachers have always divided and written the constraint off to thr side so i dunno
  10. Jul 3, 2010 #9


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    Well, most of the time you can divide. When can't you divide? What happens to everything else in that particular case?
  11. Jul 3, 2010 #10
    just say that your answer is right when h is not zero and then put a zero in place of h and solve the system again and say that this is the answer when h is zero - problem solved ! :)
    *you may need to do the same for k...

    that is also consistent with the question: determine when "such and such", so you need to check different options from h and k !
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