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Consider the following set $F=\{F^0, F^1, F^2, \ldots\}$. This set consists of positive integers which satisfy the following properties:

  1. $F^0= F^1=1$

    AND

  2. $F^n= F^{n-1} + F^{n-2}$ for all positive integers $n\geq2$.

Prove that for all positive integers $n$, the elements of the set $F$ satisfy the following identity:

$$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{n+1} = \begin{vmatrix} F^{n+1} & F^n \\ F^n & F^{n-1} \end{vmatrix}$$

where $\begin{vmatrix}\end{vmatrix}$ denotes the determinant.

Andrés E. Caicedo
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rajib
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    What have you tried? Proof by induction on $n$ strikes me as the method you should use, together with the fact that if $A$ is a matrix then $\det(A)^n = \det(A)^n$. – Clive Newstead Apr 25 '14 at 21:59
  • You could read about the Fibonacci numbers – Ross Millikan Apr 25 '14 at 22:01
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    Looks like a fancy way of saying $F_{n+1}F_{n-1}-F_n^2=(-1)^{n+1}$, which can be proved by induction. A nicer way is to work with the matrix on the left. – André Nicolas Apr 25 '14 at 22:04
  • A tip for the future. The original question, as seen rendered, was unreadable. It could only be seen properly from reading the source. What you type into the box isn't what comes out. Though LaTeX makes things easier to read, it should still be possible to see what your question is if you don't use it. – RandomUser Apr 25 '14 at 22:07
  • Hint: the identity doesn't hold only for the determinant, but also for the matrices themselves. Use induction. – egreg Apr 25 '14 at 22:16
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    A matrix multiplication like this is just another way to write a linear recurrence relation :) You can see this by writing down the result of the matrix multiplication. – Ruben Apr 25 '14 at 22:33

1 Answers1

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Hint: $$\begin{pmatrix} F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\begin{pmatrix} 1&1\\1&0 \end{pmatrix}=\begin{pmatrix} F_{n+1}+F_n&F_{n+1}\\F_{n}+F_{n-1}&F_n\end{pmatrix} $$

André Nicolas
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