Mastering Matrices: How Many Types Exist for an Order of mn (m & n Variables)?

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The discussion centers on determining the types of matrices possible for an order of mn, where m and n are variables and mn is not prime. Participants emphasize the importance of factoring mn into two factors to identify different matrix configurations. Examples like the number 6 illustrate how various combinations, such as 6x1, 1x6, 2x3, and 3x2, represent different matrix types. The conversation highlights the need for a general formula based on the unique prime factorization theorem to count the possible matrix orders. Ultimately, the focus is on understanding how to calculate the number of ways to factor mn into two factors for non-prime values.
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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)...

pleasez try
 
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electron said:
pleasez try
Who? Us?

electron said:

The Attempt at a Solution

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.
 
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.
 
Well, look at some examples first. If the order is not a prime but is, say, 6, what are the possiblities?
 
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 I've already discussed.
 
I see you love text messages...

I think you answered your own question in your last post.
 
no i didnt
 
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.
 
Do you understand that you want to count the number of ways a number can be factored into two factors?
 
  • #10
guyz..ur confusing me
 
  • #11
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.
 
  • #12
well i know that part...but incase of m*n..any idea wat can be the answer
 
  • #13
im not sure...but is there any use of permuations/combinations...
 
  • #14
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
m = m_1^{a_1} m_2^{a_2} \cdots m_p^{a_p},
where all the m_i are distinct prime numbers (if you want, ordered as m_1 < m_2 < \ldots < m_p) and the a_i 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 p = 1, m_1 = m, a_1 = 1. If m = 6 then p = 2, m_1 = 2, m_2 = 3, a_1 = a_2 = 1.
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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