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Matrix/Inconsistency Problem.

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine the value(s) of k for which the system

    -2 1 -1 x 1
    4 2 k . y = -4
    k -1 1 z 2

    (please excuse my formatting)

    2. Relevant equations



    3. The attempt at a solution

    On my first attempt I tried making an augmented matrix and row-reducing it, but it started to get really hairy and taking too long, (this is an exam question), so I figured there must be an easier
    way of solving it.

    One thing I noticed is that if I set k=2 then the first and third rows of the matrix are multiples of each other (r1=-r3), however the corresponding values in the result vector aren't, this would lead me to suspect that k=2 gives me an inconsistent system.

    IF I set k=-2 I notice the 2nd and 3rd rows are scalar multiples of each other (including the result). So I guess this is a case of linear dependence somewhere. In other words if I row-reduced the augmented matrix I'd have a row of zeros and case of one of the x/y/z being a multiple of another.

    Then I guess that leaves all other values of k to mean 1 solution

    Am I approaching this correctly? Have I just solved this? I have a test on Thursday and I'd appreciate any help or tips for looking at matrices and finding clues and short-cuts to solving them.

    If there's any experts on Linear Algebra reading, how would you approach this question? I'd love to know

    Regards
    Noel
     
  2. jcsd
  3. May 20, 2014 #2

    Zondrina

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    Usually I find it easiest to first row reduce the augmented matrix to RREF.

    You can then determine the linear dependance by testing values of ##k## as it will be clear to see which values cause an inconsistent system.
     
  4. May 20, 2014 #3

    haruspex

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    If the matrix is nonsingular, can the equations have no solution?
    If it is singular, what equation can you obtain for k? How many solutions might that have?
     
  5. May 20, 2014 #4

    D H

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    We have MathJax at this site. Note how nice this looks:
    [tex]
    \begin{bmatrix} -2 & 1 & -1 \\ 4 & 2 & k \\ k & -1 & 1 \end{bmatrix}
    \begin{bmatrix} x \\ y \\ z \end{bmatrix}
    =
    \begin{bmatrix} 1 \\ -4 \\ 2 \end{bmatrix}[/tex]

    What led you to those two particular numbers, and to the final statement? Everything is correct, but your writing makes it appear you are guessing rather than knowing.


    This is a 3x3 matrix, so it's not that hard to compute the determinant by hand. You should get ##k^2-4##. This means the only suspect values are -2 and 2. Everything other value of k yields a non-zero determinant, which in turn means the matrix is invertible. What does that mean?

    That k=2 means the first and third rows are multiples of one another but the corresponding results are not should not lead you to suspect that k=2 is inconsistent. It tells you that k=2 is inconsistent. No guessing involved!

    That k=-2 means the second and third rows are multiples of one another, including the values, means you can toss one of those rows. Is the first row linearly independent of the third (or second)? If so, this is an under-specified system rather than an inconsistent system.
     
  6. May 21, 2014 #5
    Thank you,Well it all comes with experience.
    And what exactly do you mean by an under-specified system? Is this a system with free variable(s)? Also when the det is equal to zero does this always mean either no solutions or infinite solutions but never 1 solution?
     
  7. May 21, 2014 #6

    haruspex

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    Yes. Instead of one solution there will be a subspace of solutions.
    Yes. If the det is zero then it means the LHS of one or more equations can be constructed as a linear combination of the others.
    If the same linear combination also generates the corresponding values on the RHS then those equations are redundant. If throwing them away means you now have fewer equations than variables the system is underspecified.
    But if the linear combination does not generate the corresponding values on the RHS then the equations conflict, so no solutions.
     
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