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Linear algebra question about matrices

  1. Mar 1, 2009 #1
    How would one determine the values of k such that the following matrix is the augmented matrix of a system with infinitely many solutions:

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

    Also, how would I get all values of k such that the matrix for the same to be an augmented matrix of a system with no solutions?

    It confuses me on how to find all possible values of k!

    Thank you!:smile:
     
  2. jcsd
  3. Mar 1, 2009 #2

    HallsofIvy

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    Try to row reduce it!
    Reducing it to upper triangular form, so that you have only 0s below the main diagonal, you wind up with 0 0 f(k) | g(k). There is not a unique solution if f(k)= 0. There are an infinite number of solutions if g(k)= 0 also. (If f(k)= 0 and g(k) is not 0, then there is no solution.)
     
    Last edited: Mar 1, 2009
  4. Mar 1, 2009 #3
    I would have set the determinant equal to zero. Of course the determinate can be calculated via row reduction.
     
  5. Mar 2, 2009 #4

    HallsofIvy

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    The problem with that is that if the determinant is 0, then the matrix equation may have infinitely many solutions or NO solution.
    the question here was specifically to determine k so the equation has infinitely many solutions. You need to include the right hand side to determine that.
     
  6. Mar 2, 2009 #5
    You can always substitute the roots obtained by the determinate back into the original system and then do row reduction. You can also put the system into upper triangular form while computing the determinate.
     
  7. Mar 2, 2009 #6
    Ok...

    Firstly, thank you everyone for responding...

    Secondly, I am still not getting how to determine all the values of K, so that the matrix (call it A) is the augmented matrix of a system with infinitely many solutions. How would I find all values of K?

    I understand that I need to row echelon it...but I think the way I am doing it is wrong...since the first entry cannot be 1...I start by doing Row 2 minus Row 1...and I I continue...however I am confused on what to do next in order to find k!

    Apparently I should also be able to determine all values of k so that the matrix has a system with no solutions...

    Thanks again:smile:
     
  8. Mar 2, 2009 #7
    Before you do your subtraction multiply rows one and three by (K+3) and row 2 by (k+1)
     
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