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Help in matrices

  1. Oct 1, 2007 #1
    Hi..

    I came across a small problem in matrices...wud be gr8 if some cud help.

    the problem is..

    there is matrix of order mn ( m and n are variables)

    how many types of matrices with this order are possible..keeping in mind that mn is not PRIME..(cos tat will make the answer 2)...

    plzz try
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2007 #2

    CompuChip

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    Who? Us?

    Why do you post the headings if you do not use them?

    In other words, we'd love to help, but please show us some work first.
    First you said, if m and n are prime, there are only two solutions. Why?
    Then you noted the problem if m, n are not prime.
    Now try to expand on that to find a solution.
     
  4. Oct 1, 2007 #3
    well..i said n*m shouldn't be prime because that will give us some no. like 7,19,etc.. so it has to be either row or column matrix.

    and about showing some work..sorry but i coudnt figure out where to start.
     
  5. Oct 1, 2007 #4

    HallsofIvy

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    Well, look at some examples first. If the order is not a prime but is, say, 6, what are the possiblities?
     
  6. Oct 1, 2007 #5
    incase of 6..will be something like 6x1 , 1x6 , 2x3, 3x2 ...i guess teze are all wat we have..
    but my question over here is if we have a matrix of order mxn.then wat are the no. of orders possible.

    i mentioned m*n not prime becos making it prime wud give the answer 2 ..as ive already discussed.
     
  7. Oct 1, 2007 #6
    I see you love text messages...

    I think you answered your own question in your last post.
     
  8. Oct 1, 2007 #7
    no i didnt
     
  9. Oct 2, 2007 #8
    mn can be anything..(except prime)...these are variables...so i need a general formula or some way to get the no. of orders possible.
     
  10. Oct 2, 2007 #9

    HallsofIvy

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    Do you understand that you want to count the number of ways a number can be factored into two factors?
     
  11. Oct 2, 2007 #10
    guyz..ur confusing me
     
  12. Oct 2, 2007 #11

    HallsofIvy

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    I said: Do you understand that you want to count the number of ways a number can be factored into two factors?
    If p were a prime, it can only be factored as p*1 and 1*p: that is what "will make the answer 2". If the number is 6= 2*3, it can be factored as 6*1, 1*6, 2*3, and 3*2. Those are the different "kinds" of matrices you can write of order 6.
     
  13. Oct 6, 2007 #12
    well i know that part...but incase of m*n..any idea wat can be the answer
     
  14. Oct 6, 2007 #13
    im not sure...but is there any use of permuations/combinations...
     
  15. Oct 6, 2007 #14

    CompuChip

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    Yes, that is what you will get when you formalize the conclusion you drew by trying out a couple of numbers.
    Recall the unique prime factorization theorem, by which we can write any integer as
    [tex]m = m_1^{a_1} m_2^{a_2} \cdots m_p^{a_p}[/tex],
    where all the [itex]m_i[/itex] are distinct prime numbers (if you want, ordered as [itex]m_1 < m_2 < \ldots < m_p[/itex]) and the [itex]a_i[/itex] counts how many of that factor are in m.
    This is a generalization of what you already concluded above; that is: if m is itself prime then [itex]p = 1, m_1 = m, a_1 = 1[/itex]. If m = 6 then [itex]p = 2, m_1 = 2, m_2 = 3, a_1 = a_2 = 1[/itex].
    Now multiply m by n. You will get a bunch of prime factors. In how many ways can they be recombined?
     
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