Preamble
--------

It is well known that since: 
$$
\begin{pmatrix}
 F\_{n+1} \\
 F\_n \\
\end{pmatrix} = 
\begin{pmatrix}
 1 & 1 \\
 1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
 F\_n & F\_{n-1} \\
\end{pmatrix}
$$

it is valid that:

$$
\begin{pmatrix}
 1 & 1 \\
 1 & 0 \\
\end{pmatrix}^n = 
\begin{pmatrix}
 F\_{n+1} & F\_n \\
 F\_n & F\_{n-1} \\
\end{pmatrix}
$$

By calculating determinants, it immediately follows that:

$$F\_{n-1}F\_{n+1} - {F\_n}^2
=\det\left[\begin{matrix}F\_{n+1}&F\_n\\F\_n&F\_{n-1}\end{matrix}\right]
=\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n
=\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n
=(-1)^n$$

or

$$F\_{n-1}F\_{n+1} - {F\_n}^2=(-1)^n$$

This is known as Cassini's identity.

I find both Cassini's identity and this proof exceptionally attractive.

Now, there are other identities that resemble Cassini's identity:

$$F\_{n-2}F\_{n-1}F\_{n+3} - {F\_n}^3=(-1)^nF\_{n-3}$$

$$F\_{n+2}F\_{n+1}F\_{n-3} - {F\_n}^3=(-1)^nF\_{n+3}$$

$${F\_{n-3}}{F\_{n+1}}^2-{F\_{n-2}}^2{F\_{n+3}}=4 (-1)^n{F\_{n}}$$

$${F\_{n-1}}^2{F\_{n+1}}^2-{F\_{n-2}}^2{F\_{n+2}}^2=4 (-1)^n{F\_{n}}^2$$

$$F\_{n-2}F\_{n-1}F\_{n+1}F\_{n+2} - {F\_n}^4=-1$$

Question
--------

> 
> *Is there a proof of five identities above that relies on matrices, the proof that would be attractive and similar to the mentioned proof
>  of Cassini's identity?*
> 
> 
> *Also, is there any other Fibonacci identity that follows from a suitable correspondant matrix equation?*
> 
> 
> 

Trivia
------

Not directly related to any regular (meaning having the form of an equality) Fibonacci identity, I found also some fairly surprising matrix identities involving Fibonacci numbers, and among them the strangest is:

$$\pmatrix{
3&6&-3&-1\\
1&0&0&0\\
0&1&0&0\\
0&0&1&0
}^n\pmatrix{
2197&512&125&27\\
512&125&27&8\\
125&27&8&1\\
27&8&1&1
}=\pmatrix{
F\_{n+7}^3&F\_{n+6}^3&F\_{n+5}^3&F\_{n+4}^3\\
F\_{n+6}^3&F\_{n+5}^3&F\_{n+4}^3&F\_{n+3}^3\\
F\_{n+5}^3&F\_{n+4}^3&F\_{n+3}^3&F\_{n+2}^3\\
F\_{n+4}^3&F\_{n+3}^3&F\_{n+2}^3&F\_{n+1}^3\\
}
$$

Here, one more unusual fact: $8$, $27$, and $125$ are also cubes of a Fibonacci number.