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Find values for k in a system of equations

  1. Jun 28, 2011 #1
    Hi, the following linear algebra question is causing trouble for me:

    For which values of k does the following system of equations have a) a unique solution, b) infinite solutions, and c) no solutions?

    [ 1 0 3 | 0 ]
    [ 0 1 1 | 1 ]
    [ -1 1 k | k ]

    I know that having a unique solution implies a linear independence, infinite solutions implies a dependence with consistency, and no solutions implies no consistency.

    I have tried getting this matrix into REF, but this didn't help me get any closer to the solution. Any pointers as to where I should start?
  2. jcsd
  3. Jun 28, 2011 #2


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    What is the resulting matrix? (Or the closest matrix that you can get)
  4. Jun 28, 2011 #3
    Using the following steps:
    R3=R3+R1, and
    R3=R3-R2, I get:

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

    Unless we have to get it in this form?

    [ 1 0 3 0
    [ 0 1 3+k k
    [ 0 0 -(2+k) 1-k
  5. Jun 28, 2011 #4


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    Ok, let's break this up into different cases:

    a) In the third column, third row, what operation do we perform to obtain a 1? (assuming k+1 != 0)

    b) What would (2+k) and (k-1) have to equal in order to get infinite solutions? Is this possible?

    c) Think about this one. What would (2+k) (and consequently (k-1)) have to be equal to in order to have no solution? After you have the equation set up, what do you get for k?
  6. Jun 28, 2011 #5
    a) To get 1, I'm guessing we'd have to divide by (2+k)? Since (2+k)/(2+k)=1. However, this operation would have to carry through the whole row, right?

    b) I'm not sure at all. They cant each equal zero, because that would yield k=-2 and k=1, which would turn it into an inconsistent system.

    c) Would it have to be equal to (k-1)? If so, then all I can get from the equation is k = 3+k. This is also an inconsistency, which would indicate no solutions?
  7. Jun 28, 2011 #6


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    Correct. That is the next step.

    You're partially correct. You want to set them both to 0. A row of 0's is the only way to get infinite solutions (You can verifying this by plugging in 0 for both and reducing the matrix even more) However, as you said, it's impossible to set both to 0, since you get k = -2 and k = 1. Therefore, there does not exist a k that gives infinite solutions.

    No. For there to be an inconsistent system, you would need a system that's impossible to solve. For instance:

    1 2 3 4
    5 6 7 8
    0 0 0 1

    is an inconsistent system. For the third row, this essentially says that 0x1 + 0x2 + 0x3 = 1, which is impossible. Looking at your problem now, what would (2+k) need to equal? Solve for k. What does (k-1) equal? Is this inconsistent?
  8. Jun 28, 2011 #7
    Allright, thank you so much for your help!
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