1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    gabbagabbahey

    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

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

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Matrix -show using induction
  1. Matrix induction (Replies: 4)

Loading...