Let $A=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right]$  Then $A^{2025}-A^{2020}$ is equal to 

$A=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right]$

$A^2=A\cdot A=\left[\begin{array}{lll}1& 0& 0\\ 1& 1& 1\\ 1& 0& 0\end{array}\right]$

$A^3=A^{2}A=\left[\begin{array}{lll}1& 0& 0\\ 2& 1& 1\\ 1& 0& 0\end{array}\right]$

$A^n=\left[\begin{array}{ccc}1& 0& 0\\ n-1& 1& 1\\ 1& 0& 0\end{array}\right]$

$A^{2025}-A^{2020}=\left[\begin{array}{ccc}1& 0& 0\\ 2024& 1& 1\\ 1& 0& 0\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 2019& 1& 1\\ 1& 0& 0\end{array}\right]$

$=\left[\begin{array}{lll}0& 0& 0\\ 5& 0& 0\\ 0& 0& 0\end{array}\right]$

By option verification $A^6-A.$