## anonymous 4 years ago Prove Cassini's identity using the matrix representation of the Fibonacci sequence

1. anonymous

oh is that all? easy! one question, whats cassini's identity? To google!

2. anonymous

Heh, I can open questions all over the shop now.....:-)

3. anonymous

4. anonymous

out of interest. whats the formula for the fibonacci sequence

5. anonymous

Where's the matrix?

6. anonymous

$F_n = F_{n-2} + F_{n-1}$ ?

7. anonymous

oh sorry yes your right there is none there

8. anonymous

You can represent the Fibonacci sequence as: $\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n$

9. anonymous

This is fairly easy to prove by induction

10. anonymous

define F_0 as 0, F_1 as 1, and F_2 as 1

11. anonymous

If the first equation is true, then$\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$: $= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$ $= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n-1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]$

12. anonymous

So: $\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+$

13. anonymous

now, take the determinant of both sides: $F_{n+1}F_{n-1}-F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n$

14. anonymous

The determinant of the products of square matrices is the products of the determinants of the matrices: $= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n$

15. anonymous

=$(0-1)^n=-1^n$