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Finding values of k in a system of linear equations.

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data
    For what value(s) of k does the system of linear equations below have a unique solution?

    2x+(k-1)y = 6
    3x+(2k+1)y= 9


    2. Relevant equations



    3. The attempt at a solution
    I'm uncertain how to even go about tackling this problem. Given how all the other problems beforehand were dealing with matrices, I suppose I should start there, however the variable k throws me for a loop.

    2x+ky-y = 6
    3x+2ky+y = 9

    [2 y -1 6] 1/2R1
    [3 2y 1 9]

    [1 y/2 -1/2 3] -3R1+R2
    [3 2y 1 9]

    [1 y/2 -1/2 3] 2R2
    [0 y/2 5/2 0]

    [1 y/2 -1/2 3]
    [0 y 5 0] 1/yR2

    [1 y/2 -1/2 3]
    [0 1 5/y 0]

    That is as far as I get in my blind attempt before I get stuck and no longer have any idea what I am doing. :) Any help would be greatly appreciated.
     
  2. jcsd
  3. Jan 22, 2013 #2

    SteamKing

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    Try re-writing the two equations in matrix form. Remember to include k in the matrix.
    What must be true about the matrix equations in order to have a unique solution?
     
  4. Jan 22, 2013 #3
    Alright, so I will assume from your response that my matrix was completely wrong. I am uncertain what you mean by "Try re-writing the two equations in matrix form." in that case, isn't that what I did? Should I not distribute beforehand?
     
  5. Jan 22, 2013 #4
    Also, I found this in my textbook. a. If the number of equations is greater than or equal to the number of variables
    in a linear system, then one of the following is true:
    i. The system has no solution.
    ii. The system has exactly one solution.
    iii. The system has infinitely many solutions.
    b. If there are fewer equations than variables in a linear system, then the system
    either has no solution or it has infinitely many solutions.
     
  6. Jan 22, 2013 #5

    Dick

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    The two rows of your matrix are [2,(k-1),6] and [3,(2k+1),9]. There is no y in the matrix and there are three columns, not four.
     
  7. Jan 22, 2013 #6

    Dick

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    You have two equations and two unknowns. k isn't really an 'unknown'. You just want to know how it's value affects the possibility for finding x and y.
     
  8. Jan 22, 2013 #7

    Mark44

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    Yes, your matrix was wrong.
    Below, there is no gain in expanding the terms with k.
    What you have below is what is wrong. The coefficient of y in the first row is (k - 1), and similar for the 2nd row.
     
  9. Jan 22, 2013 #8
    Alright, let's see how badly I can get this wrong.

    [2 (k-1) 6]
    [3 (2k+1) 9] 1/2R1

    [1 (k-1)/2 3]
    [3 (2k+10 9] -3R1+R2

    [1 (k-1)/2 3]
    [0 (k+5)/2 0] 2/(k+5)R2

    [1 (k-1)/2 3]
    [0 1 0]
     
  10. Jan 22, 2013 #9

    Mark44

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    What if k = -5?

     
  11. Jan 22, 2013 #10

    SteamKing

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    You are just interested in the matrix of coefficients.

    Your equations have the form:

    [A] * (x,y)^T = (6,9)^T

    The RHS is irrelevant to determining the value of k which permits a unique solution to the system.
     
  12. Jan 22, 2013 #11

    Dick

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    That's pretty good. But notice you can't divide by (k+5) if k=(-5). What do you say about the number of solutions if k=(-5) vs the number of solutions if k is not equal to -5?
     
  13. Jan 22, 2013 #12
    Well, I suppose if k = -5 then it would be undefined, whereas if it is not equal to -5 there could be infinite solutions? So, where does that leave me? There are no values for which this system of linear equations has a unique solution?
     
  14. Jan 22, 2013 #13

    Mark44

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    See if you can write an explanation with using "it" - I don't know what you're referring to.

    If k = -5, your matrix looks like this:

    $$ \begin{bmatrix} 1 & -3 & | & 3 \\ 0 & 0 & | & 0\end{bmatrix}$$

    How many solutions do you have with this augmented matrix?

    What's the situation if k ≠ -5?
     
  15. Jan 22, 2013 #14
    The "it" I was referring to was the division if k = -5 that Dick was pointing out.
     
  16. Jan 22, 2013 #15
    Also, I'm really uncertain what you were attempting to demonstrate with my second quote.
     
  17. Jan 22, 2013 #16
    I realize I may seem impossible, for this I apologize, however I am attempting to figure this all out by myself with no aid from a professor. Please bear with me.
     
  18. Jan 22, 2013 #17
    For any value of k ≠ 5, the second equation will not equal 0.
     
  19. Jan 22, 2013 #18

    Mark44

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    No worries, we are bearing with you. Can you answer the questions I asked in post #13?
     
  20. Jan 22, 2013 #19
    For the augmented matrix you wrote I can see plenty of solutions (if it indeed it equates to x-3y=3). For instance, 6-3(1)= 3, or 3-3(0)= 3, 9-3(2)= 3, etc...
     
  21. Jan 22, 2013 #20

    Mark44

    Staff: Mentor

    That's when k = -5. What happens if k ≠ -5?
     
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