matrix -show using induction

[gtfitzpatrick]

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

[gabbagabbahey]

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)

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:

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

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

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

[gtfitzpatrick]

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.