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Matrix -show using induction

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    B =
    (3 -1)
    (1 1)

    show using induction B^n = 2^(n-1)
    (2+n -n)
    (n 2-n)

    3. The attempt at a solution
    First i prove the base case so let n=1 B=2^0
    (3 -1)
    (1 1) so it holds true.

    then assume it holds true for n=k prove it true for n=k+1
    so i sub in and get
    b^k+1 = 2^k
    (3+k -k-1)
    (k+1 1-k)
    but i'm not sure where to go from here, thanks for looking! (sorry i dont know how to put in a matrix)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 6, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    You seem to be assuming that it is true for n=k+1; but you are supposed to prove that, not assume it.

    Instead assume that it is true for n=k:

    [tex]\implies B^k=2^{k-1} \begin{pmatrix}2+k & -k \\ k & 2-k\end{pmatrix}[/tex]

    Then use the fact that [itex]B^{k+1}=B^kB[/itex] to compute [itex]B^{k+1}[/itex] and show that you get

    [tex]2^{(k+1)-1} \begin{pmatrix}2+(k+1) & -(k+1) \\ (k+1) & 2-(k+1)\end{pmatrix}[/tex]
  4. Feb 6, 2009 #3
    yes, when i started to tidy up i got

    2^{(k-1)} \begin{pmatrix}2k+6) & -2k-2 \\ 2k+2 & -2k+2\end{pmatrix}

    which further tidies to your equation. this is proved because there is now k+1 where there was k's so it holds true for k+1,k+2...and all values of K...or n.
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