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Linear Algebra problem using Row Operations

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the values of k for which the system of linear equations has (i) a unique solution, (ii) no solution, (iii) infinitely many solutions. Write down the complete solution in cases (i) and (iii):
    x + y + kz = 1
    x + ky + z = 1
    kx + y + z = 1

    2. Relevant equations

    Tried sticking it into a matrix:

    [1 1 k | 1]
    [1 k 1 | 1]
    [k 1 1 | 1]

    3. The attempt at a solution

    I'm aware I then need to get that matrix into some sort of Reduced Echelon Form so I can analyse things from it. This is what I'm finding difficult as the best I can ever get is:

    [(k+2) (k+2) (k+2) | 3]
    [ 0 (k-1) (k-1) | 0]
    [ 1 1 k |1]

    That's not even close to be honest.

    I can already see that k=1 is when there's an infinite solution (as is pretty obvious) but even that I can't actually prove.

    Any guidance would be a appreciated, thank you :).
     
  2. jcsd
  3. Feb 5, 2012 #2

    Deveno

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    row-reduction isn't a bad idea, but the matrix you have isn't row-reduced. for example, you don't have a leading 1 in the first row, and you should.

    you're going to have to divide by stuff, and of course, since your denominators will contain expressions in k, there's always the chance that when these expressions = 0, your system will "blow up" so be sure to check these cases separately.
     
  4. Feb 5, 2012 #3

    Ray Vickson

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    As Deveno has indicated, you need to separate k = 0 from k ≠ 0, and do them separately. For the case k≠0 you can start with the easier system with augmented matrix
    [[k 1 1 1],[k k^2 k k],[k,k k^2 k]], where this means [row1,row2,row3]. After subtracting row 1 from the others you will have [[k 1 1],[0 k^2-1 k-1 k-1],[0 k-1 k^2-1 k-1]]. Now you need to separate k = 1 and k ≠ 1 before continuing, if you need to continue at all. I disagree with the supposed requirement of getting a reduced echelon form: that is not necessary for answering the question; it is perfectly OK to have k instead of 1 in the top left, provided that k is not 0.

    RGV
     
  5. Feb 5, 2012 #4

    Deveno

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    to be fair, getting upper (or lower) triangular is good enough. at least then, we can "back-substitute".

    getting to rref just makes the solution "transparent". i suppose it's a matter of what works for you.
     
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