anonymous
  • anonymous
Prove Cassini's identity using the matrix representation of the Fibonacci sequence
Linear Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
oh is that all? easy! one question, whats cassini's identity? To google!
anonymous
  • anonymous
Heh, I can open questions all over the shop now.....:-)
anonymous
  • anonymous
http://planetmath.org/ProofofCassinisIdentity.html come across this which answeres your question, but certainly doesnt answer mine.

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anonymous
  • anonymous
out of interest. whats the formula for the fibonacci sequence
anonymous
  • anonymous
Where's the matrix?
anonymous
  • anonymous
\[F_n = F_{n-2} + F_{n-1}\] ?
anonymous
  • anonymous
oh sorry yes your right there is none there
anonymous
  • anonymous
You can represent the Fibonacci sequence as: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]
anonymous
  • anonymous
This is fairly easy to prove by induction
anonymous
  • anonymous
define F_0 as 0, F_1 as 1, and F_2 as 1
anonymous
  • anonymous
If the first equation is true, then\[\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\]: \[= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\] \[= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n-1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]\]
anonymous
  • anonymous
So: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+\]
anonymous
  • anonymous
now, take the determinant of both sides: \[F_{n+1}F_{n-1}-F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]
anonymous
  • anonymous
The determinant of the products of square matrices is the products of the determinants of the matrices: \[= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n\]
anonymous
  • anonymous
=\[(0-1)^n=-1^n\]

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