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.