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Determinate of a Matrix

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    In my other post I figured out that when
    (2 0)
    (0 2) = m

    (2^n 0)
    (0 2^n) = m^n

    Now I have to find the determinate of each matrix M^N with powers from 1-5, 10 and 20.



    2. Relevant equations



    3. The attempt at a solution
    det (M) is 4 which is also 2^2
    det (M^2) is 16 which is also 2^4
    det (M^3) is 64 which is also 2^6
    det (M^4) is 256 which is also 2^8
    det (M^5) is 1024 which is also 2^10
    det (M^10) is 1048576 which is also 2^20

    another way of writing this would be
    det(M) the answer is 4 or 2^(2)
    det(M^2) the answer is 16 or 2^(2+2)
    det(M^3) the answer is 64 or 2^(2+4)
    det(M^4) the answer is 256 or 2^(2+6)
    det(M^5) the answer is 1024 or 2^(2+8)
    det(M^10) the answer is 1048576 or 2^(2+18)

    So basically it starts with det (M) being 4 and then each time you increase n by one the power of the answer goes up by two? How do I "generalize the pattern into an expression for det(M^n) in terms of n?
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Hi agary12! Thanks for the PM :smile:
    oooh … stop treating it like a "brainteaser" problem …

    you have the pattern …
    … so what is the determinant
    |2^n 0|
    |0 2^n| ? :smile:
     
  4. Mar 11, 2009 #3
    |2^n 0|
    |0 2^n|

    So the determinate for this is
    (2^n) * (2^N) - 0 = 2^(n+n)

    is the pattern

    |2^(n+n) 0|
    |0 2^(n+n)|
     
  5. Mar 11, 2009 #4

    tiny-tim

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    simplify!

    grrr! why do you do this?

    you had 0n before, instead of just 0, and now you have n+n …

    it's 2n !!!

    apart from that, yes, fine, the determinant is 22n ! :smile:
     
  6. Mar 11, 2009 #5
    So what would the final det(m^n) generalization look like?

    This is what I was trying to say:
    (2^n 0^n)
    (0^n 2^n)
    The above matrix shows the expression for the matrix M in terms of n. Now I was saying that to find the determinate of that expression you have to do this:
    (2^n * 2^n) - (0^n * 0^n) = det(m^n)
    So first I multiply the 2^n by the other 2^n which is (2*2)^(n+n). The 0*0=0 so the det(M^n) would equal 4^2n (whoops, I forgot to multiply the 2 by 2 and just added the exponents).

    So:
    det(m^n)= 4^2n

    Sorry that I keep doing weird things with the math, I'm just used to them trying to trick me with problems.
     
  7. Mar 11, 2009 #6

    tiny-tim

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    oooh, you made it so complicated!

    det(mn)= 22n = 4n
    so you like to get in first? :biggrin:
     
  8. Mar 11, 2009 #7
    Haha, I didn't even realize I did that first thing you mentioned. Once again, I appreciate your time and effort to help me.

    If I'm trying to do the same things I've done on the last 2 problems with a new matrix it changes the pattern and makes the "patters a bit more complex"

    They give me
    (3 1)
    (1 3)

    (3 1)^2 (10 6)
    (1 3) = (6 10)

    (3 1)^3 (36 28)
    (1 3) = (28 36)

    (3 1)^4 (136 120)
    (1 3) = (120 136)

    (3 1)^5 (528 496)
    (1 3) = (496 528)

    (3 1)^10 (524800 523766)
    (1 3) = (523766 524800)

    (3 1)^20 (5.497563382*10^11 5.497552896*10^11)
    (1 3) = (5.497552896*10^11 5.497563382*10^11)

    (3 1)^50 (6.338253001 * 10^29 6.338253001 * 10^29)
    (1 3) = (6.338253001 * 10^29 6.338253001 * 10^29)

    I don't think I can do what I did last time, that is:
    (3^n 1^n)
    (1^n 3^n)

    Can you help me discover the pattern?
     
    Last edited: Mar 11, 2009
  9. Mar 11, 2009 #8

    tiny-tim

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    ok, take the first five …

    3 10 36 136 528 …

    factor them … what do you get? :wink:
     
  10. Mar 11, 2009 #9
    3 - 1*3
    10 - 2*5
    36 - 2*2*3*3
    136 - 2*2*17
    528 - 2*2*2 *2*2*11

    This is what you meant when you said to factor them right?

    3 - (1*3)
    10 - (2*5)
    36 - (2*18), (6*6), (3*12), (4*9)
    136 - (2*68), (4 * 34), (8 * 17)
    528 - (2*264), (3*176), (4*133), (6* 88), (8*66) (11*48) (12*48), (16 *33) (22*24)

    I don't really see any patters yet, did I do something incorrectly?
     
    Last edited: Mar 11, 2009
  11. Mar 11, 2009 #10
    My friend said that this was helpful, but it seems fairly complicated to me. Do you know of any other ways to do this or is this the only way to come to the right conclusion?


    I'm confused about this part

    X= a^5+C(5,2)a^3b^2+C(5,4)ab^4 = a^5+10a^3b^2+5ab^4
    Y= C(5,1)a^4b+C(5,3)a^2b^3+b^5 = 5a^4b+10a^2b^3+b^5
    The main things that confuse me are the C(5,2), because I don't know what that means mathematically. The final X= and Y= are also confusing as I'll explain below:

    For X: If you have a^5+10a^3b^2+5ab^4 does this mean this:
    (a^5) + (10a^3b)^2+(5ab)^4
    Where do the 10 and 5 come from? How do they fit in?

    For Y:If you have 5a^4b+10a^2b^3+b^5 does it mean this:
    (5a^4b)+(10a^2b)^3 + (b^5)
    If this is correct for Y then where does the 10 come from?

    Where do the numbers 10.3 and 5.3 come from?
     
    Last edited: Mar 11, 2009
  12. Mar 11, 2009 #11
    I was just working over all of my stuff with a friend and he showed something that is extremely simple in comparison to the above method:

    [2^(2n-1) + 2^(n-1)] [2^(2n-1) - 2^(n-1)]
    [2^(2n-1) - 2^(n-1)] [2^(2n-1) + 2^(n-1)]

    I'm just wondering, where is the base 2 for everything instead of 3 or 1? How do I explain this?
     
  13. Mar 12, 2009 #12

    tiny-tim

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    hmm … you're not into this pattern thing, are you?

    let's rewrite your figures like this …

    3 - 1*3
    10 - 2*5
    36 - 4*9
    136 - 8*17
    528 - 16*33 …

    now can you see a pattern? :smile:


    If you still don't like that, here's a completely different way …

    look at the pairs …

    (1 3) (10 26) (36 28) (136 20) (528 496)

    try adding or subtracting each pair … what do you get? :wink:
     
  14. Mar 12, 2009 #13
    3 - 1*3
    10 - 2*5
    36 - 4*9
    136 - 8*17
    528 - 16*33


    In the first column it starts at one, then goes up by one, then up by two then up by four and then up by sixteen. So for the first column in the factor pairs I can see a pattern. The next factor par would be (32 * X). The second numbers are all prime. Now that I look at it again, its the same pattern on the other side as well. 5-3 = (2) 9-5=(4) 17-9=(8) 33-17=(16). So each side of the factor pair increases by 2^2, then 2^4, then 2^6 and so on?
    How do you write this mathematically?
     
  15. Mar 12, 2009 #14

    tiny-tim

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    Good, except it's 22 then 23 then 24 and so on. :smile:
    you tell us! :wink:
     
  16. Mar 12, 2009 #15
    Since I figured out how to do it for each component of the factor pair, I just don't know how I would write that for each entry of the matrix.

    would it look something like this: 2^(n+1)? For one entry that is. But the problem I have with this is the n variable. Should N represent the increasing exponents or is there a better way to do it? I'm almost certain that each Matrix entry will have 2 terms, but I don't know how to get there.

    I believe that the final answer looks like this:

    (x y)^n
    (y x)
    =
    (a b)
    (b a)


    a = [(x+y)^n + (x-y)^n] / 2
    b = [(x+y)^n - (x-y)^n] / 2

    The problem is that I can't explain it yet.
     
    Last edited: Mar 12, 2009
  17. Mar 12, 2009 #16

    tiny-tim

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    Hint: if you start with something, and add 2 then add 4 then add 8 then add 16 … and so on … and finally add 2n, what have you added altogether? :smile:

    (if you still don't get it, try it in binary)
     
  18. Mar 12, 2009 #17
    I don't know what you mean by trying it in binary. You've been adding powers of 2 right? so is it just 2^n?
     
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