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Rank of a matrix and more

  1. Sep 6, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Find the rank of the matrix A,where
    [tex]A= \left(
    \begin{array}{cccc}
    1 & 1 & 2 & 3\\
    4 & 3 & 5 & 16\\
    6 & 6 & 13 & 13\\
    14 & 12 & 23 & 45
    \end{array}
    \right)
    [/tex]

    Find vectors[tex]x_0[/tex]and[tex]e[/tex] such that any solution of the equation

    [tex]Ax= \left(
    \begin{array}{c}
    0\\
    2\\
    -1\\
    3
    \end{array}
    \right)
    [/tex] [tex](*)[/tex]
    can be expressed in the form [tex]x_0+\lambdae[/tex] where [tex]\lambda\epsilonR[/tex]

    Hence show that there is no vector which satisfies [tex]*[/tex] and has all its elements positive




    2. Relevant equations

    First attempt at such a question, so unknown are any relevant equations

    3. The attempt at a solution
    Well for the first part to get the rank I put A in RRE form and then counted the number of non-zero rows and got for so [tex]r(A)=4[/tex]

    now for the second part,I thought to solve the equation by multiplying by [tex]A^{-1}[/tex] and finding [tex]x[/tex] but then I realized that I have no idea where to get [tex]x_0[/tex] or [tex]\lambda[/tex] or [tex]e[/tex]

    can anyone show me how to do these types of questions or can show me some similar example?
     
  2. jcsd
  3. Sep 6, 2007 #2

    Hurkyl

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    Well, you made a mistake somewhere in here.

    You might have guessed that -- if you can write any solution in the form the problem asks for, what does the rank of the matrix have to be?

    (Hint: what does the nullity of the matrix have to be?)
     
  4. Sep 6, 2007 #3

    rock.freak667

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    Did I do the row-reduction wrong?
    well from wikipedia...[tex]rank(A)+Nullity(A)=n[/tex] well [tex]n=4[/tex] in this case

    BTW...This is the first time I have heard of nullity
     
  5. Sep 6, 2007 #4

    Hurkyl

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    I believe so. The statement of the problem implies the rank is not 4. (In fact, it implies a specific number for the rank) I tried once to do the row reduction myself, and I got the number I expected.
     
  6. Sep 6, 2007 #5

    rock.freak667

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    Well I believe I did it over correctly and got [tex]r(A)=3[/tex]
     
  7. Sep 6, 2007 #6
    yes, you seems to be correct, if this is what you were trying to get:
    [tex]\pmatrix{1 & 1 & 2 & 3\cr 0 & 1 & 3 & -4\cr 0 & 0 & 1 & -5\cr 0 & 0 & 0 & 0}[/tex]

    use maxima!!

    [​IMG]
     
    Last edited: Sep 6, 2007
  8. Sep 6, 2007 #7

    rock.freak667

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    But how do I use the fact that [tex]r(A)=3[/tex] and the nullity to find the vectors in that form?
     
  9. Sep 6, 2007 #8

    Hurkyl

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    Well, how do you normally solve systems of equations? Have you tried that?
     
  10. Sep 6, 2007 #9

    rock.freak667

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    Well normally for that matrix I would just augment it and try to put it in RRE form but then i dont know where [tex]x_0[/tex] and [tex]e[/tex] and [tex]\lambda[/tex] comes in
     
  11. Sep 6, 2007 #10

    Hurkyl

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    Well, try solving it first, then think about it.

    By the way, you can edit your original post to fix that one formula; you're supposed to put spaces between things. And it looks a lot nicer if you use [ itex ] instead of [ tex ] for stuff in paragraphs.
     
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