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Explanation of solution to given question (subspace)

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data



    -4x1 +2x2 =2
    4x1 -3x2 -2x3 =-3
    2x1 -x2 +(k - k2)x3 =-k

    Find the values of k for which the system has 1) unique solution, 2) infinitely many solution and 3) no solution



    3. The attempt at a solution

    In REF: the matrix is
    -4 2 0 | 2
    0 -1 -2 | -1
    0 0 k(1-k) | 1-k


    1) I do not understand why the solution sets x3 = y to give a single parameter solution set.


    2) for a system to have infinitely solution k must be 1 because 1(1-1) = 0 and 1-1 = 0
    0 = 0 implies infinitely many solution

    3) for a system to have no solution, the solution must be inconsistent
    and so, k(1-k) must equal to 0 for some value k and 1-k gives a value that is an element of real number with the exclusion of 0. For this to be true, k = 0.
    0(1-0) = 0 and 1-0 = 1
    we see that 0 = 1 but this produces a contradiction.
     
  2. jcsd
  3. May 29, 2014 #2

    ehild

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    If k=1 x3 is arbitrary. y means just an arbitrary number. And you can write the solution set in terms of y.

    Correct. But what happens if k≠1 and k≠0? What is (are) the solution(s)?


    ehild
     
  4. May 29, 2014 #3
    I'm still not capturing the full essence of part (1). Maybe some simple examples?

    part(3): if k ≠ 1 and k ≠ 0 then we can pick any values of k? But what is the corollary from this?
     
  5. May 29, 2014 #4

    ehild

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    Part (1) asks about unique solution (for a given k) K can be anything, but 1 and zero. What is the solution set in terms of k?

    Part(2) : infinite many solution exist if k=1, as x3 can be arbitrary. What are x1 and x2 in terms of x3?

    Part 3 : no solution exist when k=0.

    (1) Remember that a row of your matrix equation means an equation for x1,x2,x3. The last row of the reduced matrix means the equation k(1-k) x3=1-k. If k≠1 and k≠0 you get x3=1/k.
    The second row means equation -x2-2x3=-1. Substitute 1/k for x3: -x2-2/k=-1, x2=1-2/k.
    Continue with the first row and find x1.
    Example: k=2. x3=0.5, x2=0.

    You see that the solution set is unique for a given value of k.
    In case (2) k=1 and x3 is arbitrary. x3=y
    The second raw means -x2-2x3=-1→-x2-2y=-1→x2=1-2y.
    Do the same with the first row.
    If y=0, for example, x3=y=0, x2=1 and x1=0.

    ehild
     
    Last edited: May 29, 2014
  6. May 29, 2014 #5
    It is clear now. I was earlier able to work to the part where k(1-k)x3 = (-k+1)
    simplifying produces x3 = 1/k but was confused as to what I should do with what I had especially since it conflicted with what my answer key showed.

    In response to your question, x3 = 1/k. If x3 = y, 1/k = y.

    ∴ x3 = y

    -x2 -2x3 = -1
    -x2 -2y = -1
    ∴ x2 = 1-2y

    -4x1 +2x2 =2
    -4x1 +2(1-2y) 2
    ∴x1 =-y

    1/k can be substituted into y (preferences)
     
  7. May 29, 2014 #6

    ehild

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    No, y comes in when k=1.

    If k is neither 1 nor zero, there is a unique solution , and that is case (1)

    If k=1 you have a free parameter - y - in the solution. That is case (2)

    x3=y; x2=1-2y, x1=????


    ehild
     
  8. May 29, 2014 #7
    I should have known not to complicate the answer by setting 1/k = y.

    But I have this one.

    If k = 1, we have 1(1-1) | (-1+1) and this produces 0 = 0.
    We have an infinite solution set.
    0x3 = 0
    setting x3 = y we see that via back substituition

    x2 = 1-2y
    x1 = [2+2(1-2y)]/-4 :tongue: (lazy to simplify but I have the conceptual idea)
     
  9. May 29, 2014 #8

    ehild

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    Good! :biggrin:

    ehild
     
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