Matrices, determinants, and applications to identities involving Fibonacci numbers 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.
 A: Following statement and proof are interesting:

Prove that
$$F_{2n-1} = {F_{n}}^2 + {F_{n-1}}^2$$

Let us start with
$$
\begin{pmatrix}
   1 & 1 \\
   1 & 0 \\
\end{pmatrix}^n = 
\begin{pmatrix}
   F_{n+1} & F_n \\
   F_n & F_{n-1} \\
\end{pmatrix}
$$
If we split $n$ into $p$ and $q$, this can be rewriten as:
$$
\begin{pmatrix}
   1 & 1 \\
   1 & 0 \\
\end{pmatrix}^p
\begin{pmatrix}
   1 & 1 \\
   1 & 0 \\
\end{pmatrix}^q
= 
\begin{pmatrix}
   1 & 1 \\
   1 & 0 \\
\end{pmatrix}^{p+q}
$$
which is actually the same as:
$$
\begin{pmatrix}
   F_{p+1} & F_{p} \\
   F_{p} & F_{p-1} \\
\end{pmatrix}
\begin{pmatrix}
   F_{q+1} & F_{q} \\
   F_{q} & F_{q-1} \\
\end{pmatrix}
 = 
\begin{pmatrix}
   F_{p+q+1} & F_{p+q} \\
   F_{p+q} & F_{p+q-1} \\
\end{pmatrix}
$$
and, by calculating the right bottom element of the right hand side matrix, follows that:
$$F_{p+q-1} = F_p F_q + F_{p-1} F_{q-1}$$
Since we can assume this is valid for any natural $p$ and $q$, it will be valid for $p=q=n$ too:
$$F_{2n-1} = {F_{n}}^2 + {F_{n-1}}^2$$
and this is what we wanted to prove.

Moreover, from matrix equation, this is also valid:
$$F_{p+q} = F_{p+1}F_q + F_{p} F_{q-1}$$
and this leads to identity:
$$F_{2n} = F_{n+1}F_n + F_{n} F_{n-1}$$

There is also a possibility to split $n$ into $p$, $q$, and $r$, and obtain different identities. 
A: Let's look at your first identity.  Express everything in terms of $F_{n-1}$ and $F_n$.
$$ \eqalign{F_{n-2} F_{n-1} F_{n+3} - F_{n}^3  &= 3 F_{n}^2 F_{n-1} - F_n F_{n-1}^2 - 2 F_{n-1}^3 - F_{n}^3 \cr
&= (F_n - 2 F_{n-1})(-F_n^2 + F_n F_{n-1} + F_{n-1}^2)\cr
(-1)^n F_{n-3} &= -(-1)^n (F_n - 2 F_{n-1})\cr}$$
so the identity reduces to $F_n^2 - F_n F_{n-1} - F_{n-1}^2 = (-1)^n$
which is equivalent to Cassini.
A: The "strange matrix identity" at the bottom is just a linear recurrence among the cubed Fibonacci numbers:
$$F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$$
which looks somewhat less formidable than the matrix equation.
To derive this, we start with Binet's formula
$$ F_n = \frac{\phi^n-(-\phi^{-1})^n)}{\sqrt 5} $$
and cube it to get
$$F_n^3=\frac{(\phi^3)^n-3(-\phi)^n+3(\phi^{-1})^n-(-\phi^{-3})^n}{5\sqrt5}$$
The polynomial whose roots are the numbers raised to the $n$th power here is
$$(x−\phi^3)(x+\phi)(x−\phi^{-1})(x+\phi^{-3})=x^4−3x^3−6x^2+3x+1$$
which supplies the coefficients of the above recurrence.     
In a similar way one can derive $$F^2_{n+3}=2F_{n+2}^2+2F_{n+1}^2-F_n^2$$ and its associated matrix form
$$ \begin{pmatrix}2 & 2& -1\\1&0&0\\0&1&0\end{pmatrix}^n
\begin{pmatrix}1\\1\\0\end{pmatrix} =
\begin{pmatrix}F_{n+2}^2\\F_{n+1}^2\\F_n^2\end{pmatrix} $$
See A055870 for the corresponding coefficients for higher powers of Fibonacci numbers.
