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Matrices - finding critical values

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Find k so that the system has exactly one solution.

    [tex]\left{ \begin {array} {rcl} x - y + 2z = 1 \\ -x + y - z = 2 \\ x + ky + z = 3 \end {array} \right .[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Ok, so I create an augmented matrix here:
    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ -1 & 1 & -1 & 2 \\ 1 & k & 1 & 3 \end {array} \right][/tex]

    Here I do the following: R2+R1 and R3-R1.
    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & k+1 & -1 & 2 \end {array} \right][/tex]

    R2 <=> R3.
    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    (1/k+1)R2.
    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 1 & -1/(k+1) & 2/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    At this point, I'm stuck. How do I know which k will give the system exactly one solution?
     
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 4, 2007 #2
    Is your first equation correct? It does not match with your matrix.

    Have you learned about eigenvalues?
     
  4. Feb 4, 2007 #3
    Sorry; I just fixed it. Everything should be correct now. We haven't learned about eigenvalues yet.
     
  5. Feb 4, 2007 #4
    Would reducing your matrix all the way to the identity matrix help?
     
  6. Feb 4, 2007 #5
    We haven't even gone over the identity matrix yet. :redface: I think I may have come up with a solution:

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    OK, in this form, if k = -1, the second and third rows would be contradictory (i.e. no solution):

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    However, if k were anything else, there should be exactly one solution, right? So is the answer all real numbers except -1, or am I totally off-base here?
     
  7. Feb 4, 2007 #6
    Looks like 'a' solution. How would you know if there were others, or not?

    The identity matrix has ones on the diagonal, & zeros everywhere else. Basically continue your reduction method, or back-substitute from where you are, to obtain complete expressions for x,y,z in terms of k & then make a decision.

    This will give you the full solution for x, y, z & you could then re-check your current solution.
     
  8. Feb 4, 2007 #7
    Ok, I've reduced my matrix to the identity matrix. Here are my steps:

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & 0 & 5 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 0 & -5 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    [tex]\left[ \begin {array} {ccc|r} 1 & 0 & 0 & 5/(k+1) - 5 \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    After all these steps, my final identity matrix looks like this:
    [tex]\left[ \begin {array} {ccc|r} 1 & 0 & 0 & -5k/(k+1) \\ 0 & 1 & 0 & 5/(k+1) \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]

    So now it looks like the possible values of k are all real numbers except -1 and 0. Is this correct?
     
  9. Feb 5, 2007 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why could k not be 0? Isn't x= 0, y= 5, z= 3 an acceptable solution?

    The matrix equation, Ax= b, will have a single solution as long as det(A) is not 0. Since there is only one k in the matrix, det(A)= 0 will be a linear equation in k. There should be only one value of k for which this system does not have a single solution.

    You could have seen that back in your original reduction:
    [tex]\left[ \begin {array} {ccc|r} 1 & -1 & 2 & 1 \\ 0 & k+1 & -1 & 2 \\ 0 & 0 & 1 & 3 \end {array} \right][/tex]
    If the "pivot", k+ 1, is 0 then you cannot have a single solution.
     
  10. Feb 5, 2007 #9
    Wow, how dumb of me... :redface: Of course k=0 is a solution! For some reason I was looking at it like 1=0. :confused: So it looks like the only value that won't work is -1, as I thought earlier. Thanks!
     
  11. Feb 5, 2007 #10

    radou

    User Avatar
    Homework Helper

    Hence, it's always necessary to create a 'discussion' before dividing with k+1, as in post #7, since, for k = -1 the term is undefined.
     
    Last edited: Feb 5, 2007
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