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Infinite solution to system with no free variables?

  1. Mar 10, 2016 #1

    TheBlackAdder

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    • Member warned about posting with no template, and with incomplete problem statement
    1. The problem statement, all variables and given/known data
    The assignment is to find all values of k (in R) for which the system has 0 solutions, 1 solution and infinite solutions. If there are infinite solutions, find the amount of free variables.

    The system of linear equations:
    kx + (k+1)y + z = 0
    kx + y + (k+1)z = 0
    2kx + y + z = k+1

    2. Relevant equations
    FSLs9Mi.png

    Now k=-1
    mnW7RpP.png

    3. The attempt at a solution
    So I was correct that this system is undefined when k=0.
    Then I thought this system had infinite solution when k =/= 0. But here I'm wrong.
    Apparently if k=-1 then there are infinite solution with one free variable. However, when I plug in -1 then all solutions are -1/2.

    I don't understand why this is infinite.

    I did my row operations as practise.
    c9Qq4ss.jpg
     
    Last edited: Mar 10, 2016
  2. jcsd
  3. Mar 10, 2016 #2

    Mark44

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    The proper term is that the system is inconsistent; i.e., there are no solutions.
    At this point, you can assume that ##k \ne 0##, so continue row reduction from here. The possibilties now are that for some value(s) of k, the system has an infinite number of solutions, and for some other value(s), the system has a single solution.
    That's not what I get. If k = -1, there are an infinite number of solutions. Check your arithmetic.

    Also, it wasn't clear until well into your hand-written work that your matrix is an augmented matrix, and that the system is three equations in three unknowns. You should have included the system of equations in the problem statement of the template (which you deleted).
     
  4. Mar 10, 2016 #3

    TheBlackAdder

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    Well that's what I don't understand. In the row reduced echelon form there is only one k left. If k=-1 then you get -1/2.
    I do realise if k=-1 from the start, then you'll get a system of infinite solutions, but how do you figure out -1? I shouldn't have row reduced the matrix?
     
  5. Mar 10, 2016 #4

    haruspex

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    Check back through your working during the row reduction. Look for each place you divided through by an expression with k in it. What do you need to consider?
     
  6. Mar 10, 2016 #5

    Mark44

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    By row reduction, I get the augmented matrix down to this:
    ##\begin{bmatrix}k & k + 1 & 1 & | &0\\ 0 & k & -k & | & 0 \\ 0 & 0 & 2k^2 + 2k & | & -k^2 - k \end{bmatrix}##

    Where we're going to have trouble is in the last row. The third term in that row is 0 if k = 0 or k = -1 (which you have already said).

    Go back to the original matrix, and plug in k = 0, and row reduce this matrix. This should tell you that the system is inconsistent if k = 0. Again, using the original matrix, plug in k = -1, and row reduce this matrix. You should see that you have one free variable in this situation.
     
  7. Mar 10, 2016 #6

    Ray Vickson

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    I get something a bit different: if the equations are written as (just by reordering the variables and the equations)
    y + k*x + (k+1)*z = 0
    y + 2*k*x + z = k+1
    (k+1)*y + k*x + z = 0
    the augmented matrix is
    [tex] \left|\begin{matrix}
    y & x & z & | &\text{RHS} \\ \hline
    1 & k & k+1 &|& 0 \\
    1 & 2k & 1 &|& k+1 \\
    k+1 & k & 1 &|& 0
    \end{matrix}\right|
    [/tex]
    After row-reducing on columns 1 and 2 we have
    [tex]
    \left|\begin{matrix} y & x & z& | & \text{RHS} \\ \hline
    1 & k & k+1 &|& 0 \\
    0 & k & -k &|& k+1 \\
    0 & 0 & -2k(k+1) &|& k(k+1)
    \end{matrix} \right|
    [/tex]
    This is a row-reduced form, but not a RREF.

    Note that when ##k = 0## the second row gives an inconsistent equation ##0y+0x+0z = 1##. The cases ##k = -1## and ##k \neq -1,0## can be treated similarly, starting from the row-reduced form.
     
    Last edited: Mar 11, 2016
  8. Mar 11, 2016 #7

    TheBlackAdder

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    That I cancel out / reduce the possibilities for finding free variables?
     
  9. Mar 11, 2016 #8

    haruspex

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    No. In algebra, what is it illegal to divide by?
     
  10. Mar 11, 2016 #9

    TheBlackAdder

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    Zero. But dividing by k is not dividing by zero?

    As far as I know, rewriting the system in a different form does not change the original system? But apparently it does because when k=-1 in the first image you get a free variable and infinite solutions, whereas in the second image you get one solution. {-1/2,-1/2,-1/2}

    Okay, the question I must ask myself is why. I'll try and figure it out.
    zfRI542.png = sOAq83c.png

    PS It has been a real long time since I came into contact with math and went further than memorizing formulas. I really want to understand why things work now, so sorry for asking stupid questions.
     
    Last edited: Mar 11, 2016
  11. Mar 11, 2016 #10

    haruspex

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    It is when k is zero.
    Given an equation like x f(x)= x g(x) you should argue " either x=0 or ....".
     
  12. Mar 11, 2016 #11

    TheBlackAdder

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    I did not realise it wasn't allowed to divide by an unknown which could be zero. I did so many wrong calculations doing that.
     
  13. Mar 11, 2016 #12

    Ray Vickson

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    That is not quite right: you ARE allowed to divide by k as long as k ≠ 0. So you can split things up into cases, for example:
    Case (1): k = 0 ---> proceed.
    Case (2): k ≠ 0 ---> proceed. Dividing by k is OK in this case.
    In Case (2) you may run into further problems if k = -1, so you can have:
    Case (2a) k = -1 ---> proceed; and
    Case (2b) k ≠ -1 (and, f course, k ≠ 0). Dividing by k and k+1 are both OK in this case.
     
  14. Mar 11, 2016 #13

    TheBlackAdder

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    Sorry, I had to be more specific. I meant when there is a leading k in the matrix, just leave it as k. Don't multiply the row by 1/k just to get a one. In all my exercises I did stuff like that because Wolfram|Alpha (I use it as verification) did it also. So I applied it everywhere.

    Btw, don't you mean: Case (1): k = 0 ---> don't proceed.
     
  15. Mar 11, 2016 #14

    Ray Vickson

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    No, I meant "proceed", but not necessarily in the same way you would if k were nonzero. There can be several ways to proceed, depending on conditions.
     
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