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Rank of matrix problem

  1. Apr 1, 2008 #1
    find the value for k for which the matrix

    A=
    | 9 -1 11 |
    |-6 5 -16 |
    | 3 2 k |

    has rank= 2


    * the spacing on the matrix doesn't seem to want to stay formatted, but it's a 3X3 with row 1= (9, -1, 11), row 2= (-6, 5, -16) and row 3=(3,2, k)
    3. The attempt at a solution

    - I tried to solve this using row reduction and then solve for k, but I don't think thats right at all. Can someone please explain this problem to me and the technique for solving it?
     
    Last edited: Apr 1, 2008
  2. jcsd
  3. Apr 1, 2008 #2

    rock.freak667

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    [tex]\left(
    \begin{array}{Ccc}
    9 & -1 & 11\\
    -6 & 5 & -16 \\
    3 & 2 & k\
    \end{array}
    \right)[/tex]

    What you need to is row reduction to Row echelon form, the number of non-zero rows is the rank,2 in this case. So there should be 2 non-zero rows.
     
  4. Apr 1, 2008 #3
    - That is what I tried to do, but then I don't know how to find k. I got this:

    row 1= (3, 2, k)
    row 2= (0, 9, -16+2k)
    row 3= (0, -7, 11-3k)

    I don't know how to solve for k at this point, would it just be guess work, or is there a technique to use?
     
  5. Apr 1, 2008 #4

    rock.freak667

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    Try these operations
    [tex]3R_3-R_1,9R_2+6R1[/tex]


    that should give you something better.

    Edit: Do you know the easiest method to get to RE form when given a matrix?
     
    Last edited: Apr 1, 2008
  6. Apr 1, 2008 #5
    There are a few ways of approaching this, but this is the first that came to my mind.

    For it to be rank 2, you should try finding k such that the third row is a linear combination of the first two. So try finding c1 and c2 such that:

    c1 R1 + c2 R2 = R3.

    The first two components will give you a 2x2 system, and you can then solve for c1 and c2. Then, the third component will give you k.

    Hopefully that makes sense and is helpful!
     
  7. Apr 1, 2008 #6
    yes, that made it much easier. thanks.
     
  8. Jul 24, 2010 #7
    just do determine of matrix zero and find value of k
     
  9. Jul 24, 2010 #8

    rock.freak667

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    You did not really have to bump a 2 year old thread.
     
  10. Jan 30, 2012 #9
    Heyya Snogger
     
    Last edited: Jan 30, 2012
  11. Jan 30, 2012 #10
    A = 9 -1 11
    -6 5 -16
    3 2 k

    = taking 3 common from C1

    3 -1 11
    -2 5 -16
    1 2 k

    = R2->R2+2(R3)

    3 -1 11
    0 9 k-16
    1 2 k

    =R3->3(R3)-R1

    3 -1 11
    0 9 k-16
    0 7 3k-11

    =R2->R2/9

    3 -1 11
    0 1 k-16/9
    0 0 3k-11

    =R3->R3+7R1

    3 -1 11
    0 1 k-16
    0 0 3k-66

    Since,the rank of the matrix is 2

    3k-66=0
    3k=66
    k=22

    Sorry if I m wrong just gave a try...
     
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