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Matrix with power of n

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    a and n could be any real number, please evaluate the value.

    [tex]
    B =\left [
    \begin{array}{ccc}
    a&1&0\\
    0&a&0\\
    0&0&a

    \end{array}
    \right ]
    [/tex]n

    I've no idea how to approach this question, because i have never came across the matrix with power not equals to -1.
    could somebody give me some hints?

    I've try to let n = -1 by using gauss jordan elimination, and let then elements become 1/a, 1/a2 and etc. but this seems incorrect and not what the question asking for..
     
  2. jcsd
  3. Oct 9, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can either expand it for small values of n and guess the pattern and prove it works by induction (if you have to). Or write the matrix as aI+J, where I is the identity and J is zero except for that off diagonal 1. Now, I and J commute, so you can apply the binomial theorem to (aI+J)^n.
     
  4. Oct 9, 2009 #3

    Mark44

    Staff: Mentor

    Are you sure that the problem says that n could be any real number? You're going to have a hard time raising your matrix to the power [itex]\pi[/itex], say.
     
  5. Oct 9, 2009 #4
    yeah, the question states the a and n is any real value. but why power of pi?

    but i'm not very sure this question is asking me to find the n and a value or prove it could be done by any real value(for a and n)
     
  6. Oct 9, 2009 #5

    Mark44

    Staff: Mentor

    I understand that the problem is not asking you to find a and n, but if n can be any real number, then there is nothing that prevents n from being [itex]\pi[/itex]. If so, what does it mean to raise a matrix to an irrational power? A more reasonable interpretation, IMO, would be that n is a positive integer, for which math induction would be one approach.
     
  7. Oct 9, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm with the others. If both a and n can be any real numbers, then for some combinations of a and n, the matrix won't even exist. For example, if a= 0 and n= -1, B-1 will not exist. If a is negative and n = 1/2, you are going to run into complex numbers. If a can be any real number, and n any positive integer, the problem is relatively simple. Calculate B, B2, B3, make a good guess at what the general form will be and, as Mark44 suggested, prove it by induction.
     
  8. Oct 9, 2009 #7

    Dick

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    Science Advisor
    Homework Helper

    Boo hoo. Nobody likes my binomial theorem suggestion??
     
  9. Oct 9, 2009 #8

    Mark44

    Staff: Mentor

    I like it. I was just offering another possible slant.
     
  10. Oct 9, 2009 #9
    yeah, i like this solution =)
    thanks all
     
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