12.2. Let the matrix $A=\left(\begin{array}{cc}-a & \frac{1}{1-a} \\ a^{3}-1 & 1+a\end{array}\right)$, where $a \in \mathbb{R} \backslash\{1\}$. Determine $A^{2021}-A^{2020}$.
Solution:

We observe that

$$
A^{2}=\left(\begin{array}{cc}
-a-1 & \frac{1}{1-a} \\
a^{3}-1 & a
\end{array}\right), \quad A^{3}=\left(\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right)=-I_{2}
$$

Then $A^{2020}=\left(A^{3}\right)^{673} \cdot A=-A, A^{2021}=\left(A^{3}\right)^{673} \cdot A^{2}=-A^{2}$.

We obtain $A^{2021}-A^{2020}=-A^{2}+A=I_{2}$.