# Discrete math matrix?

• Jul 19th 2009, 02:24 AM
orendacl
Discrete math matrix?
How do I show:

Given we let A=:
|1 1|
|1 0|
That:
A^n is equal to the matrix:
|fn+1 fn|
|fn fn-1|

for whenever R is positive.
• Jul 19th 2009, 04:21 AM
running-gag
Quote:

Originally Posted by orendacl
How do I show:

Given we let A=:
|1 1|
|1 0|
That:
A^n is equal to the matrix:
|fn+1 fn|
|fn fn-1|

for whenever R is positive.

Hi

Let $\displaystyle A^n = \left(\begin{array}{cc}a_n&b_n\\c_n&d_n\end{array} \right)$

Then $\displaystyle A^{n+1} = A \cdot A^n = \left(\begin{array}{cc}a_n+c_n&b_n+d_n\\a_n&b_n\en d{array}\right)$

Therefore
$\displaystyle a_{n+1} = a_n + c_n$
$\displaystyle b_{n+1} = b_n + d_n$
$\displaystyle c_{n+1} = a_n$
$\displaystyle d_{n+1} = b_n$

The first and third relations give $\displaystyle a_{n+2} = a_{n+1} + c_{n+1} = a_{n+1} + a_n$

The second and fourth relations give $\displaystyle b_{n+2} = b_{n+1} + d_{n+1} = b_{n+1} + b_n$

Check the first terms of $\displaystyle a_n$ and $\displaystyle b_n$ to determine the index of f

The third and fourth relations give the rest of the solution
• Jul 25th 2009, 07:15 AM
orendacl
??? Here is what I have so far:
I said that the relation for 3rd and 4th should be:
cn+2 = cn+1+dn+1 = cn+1+bn

I tried everything except I do not know what to do given the relations for the problem...