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System of equations / matrices

  1. Nov 17, 2014 #1
    • Member warned about not using the homework template
    Find all values of h and k such that the system has no solutions, a unique solution, infinitely many solutions

    hx +6y =2
    x (h+1)y =2k

    I can't seem to augment the matrix. Am I allowed to multiply buy variables h / k?

    I can find the determinant: h^2 +h -6

    Then make it equal to 0 and solve; h = 3 and -2

    I'm not entirely sure how that corresponds to the question.

    Thanks a lot for any help; it's a little urgent.
    :)
     
    Last edited: Nov 17, 2014
  2. jcsd
  3. Nov 17, 2014 #2

    BruceW

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    how did you get that value of the determinant? I get something different. firstly, it is best to actually write down the matrix equation, since I'm guessing that is where you got the value of the determinant from.
     
  4. Nov 17, 2014 #3
    h,6 II 2
    1, (h+1) II 2k

    I swapped the rows when I found mine, but if you don't you get: 6 -h^2-h
    Unless I'm doing something terribly wrong.
     
  5. Nov 17, 2014 #4
    Is it possible to apply gaussian elemination?
     
  6. Nov 17, 2014 #5

    ehild

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    Yes. Do it.
     
  7. Nov 17, 2014 #6

    BruceW

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    Oh, right. I didn't see the y in the matrix when I looked at it before. Was it always there? I guess maybe I just didn't notice it. Anyway, yes I agree that you want h2+h-6 equal to zero. But are the solutions really 3 and -2 ?

    Think about it case by case. What is possible when the determinant is nonzero, and what is possible when the determinant is zero?
     
    Last edited by a moderator: Nov 17, 2014
  8. Nov 17, 2014 #7
    I don't seem to be able to. Is it possible to multiply a line by 1/h?
     
  9. Nov 17, 2014 #8

    ehild

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    It is possible if h is not zero. But you can interchange row 1 with row 2.
     
  10. Nov 17, 2014 #9
    So I have swapped the rows and made the bottom left = 0

    1, h+1 II 2k
    0, 6/h -h -1 II 2/h -2k

    Am I on the right tracks?
     
  11. Nov 17, 2014 #10

    ehild

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    Do not divide by h. Subtract h times of the first line from the second one. You get 0, 6-h(h+1)||2-2kh
     
  12. Nov 17, 2014 #11
    Should I attempt to solve 6-h(h+1) = 2-2kh?
    (I arrived at this earlier)
     
  13. Nov 17, 2014 #12

    Mark44

    Staff: Mentor

    The problem with working with matrices and augmented matrices is that it's too easy to lose sight of what the matrices represent. If your bottom row is
    ##\begin{bmatrix} 0 & h^2 + h - 6 & | & 2kh - 2 \end{bmatrix}##
    what equation does this represent?

    In light of your answer, would it be a good idea to set h2 + h - 6 equal to 2kh - 2?
     
  14. Nov 17, 2014 #13

    When the x coeff. is 0, the y coeff. (h^2 + h - 6) is = to 2kh -2?

    Probably not but I seem a little lost as to what needs to be done.
     
  15. Nov 17, 2014 #14

    Mark44

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    No. The coefficient of y is h2 + h - 6. What equation does this represent?
     
  16. Nov 17, 2014 #15
    (h^2 + h - 6)y = 2kh -2?
     
  17. Nov 17, 2014 #16

    Mark44

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    Yes. Can you always solve this equation for y?
     
  18. Nov 17, 2014 #17
    Not if h = 3 or -2?
     
  19. Nov 17, 2014 #18

    Mark44

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    OK, if h
    As was mentioned way back in post #6, the solutions are not h = 3, h = -2. What are the correct values?

    Once you get those, how does k enter into things?
     
  20. Nov 17, 2014 #19
    h=2, -3?

    Sub the values into 1, h+1 II 2k?

    Sorry about this, my head's not functioning.
     
  21. Nov 17, 2014 #20

    Mark44

    Staff: Mentor

    How? That's not an equation.
    You can't do this stuff if your head's not functioning. Maybe you should take a break and come back when it is functioning.

    This is the equation.
     
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