$\text{ Let }\mathrm{A}=\left(\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right)\text{ then }{\mathrm{A}}^{2025}-{\mathrm{A}}^{2020}=$

$$\begin{gathered} \begin{array}{l}\mathrm{A}=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]\Rightarrow {\mathrm{A}}^2=\left[\begin{array}{lll}1& 0& 0\\ 1& 1& 1\\ 1& 0& 0\end{array}\right] \\ \Rightarrow {\mathrm{A}}^3=\left[\begin{array}{lll}1& 0& 0\\ 2& 1& 1\\ 1& 0& 0\end{array}\right], \\ \Rightarrow {\mathrm{A}}^4=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 1\\ 1& 0& 0\end{array}\right] \\ \Rightarrow {\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{ccc}1& 0& 0\\ \mathrm{n}-1& 1& 1\\ 1& 0& 0\end{array}\right]\\ \therefore {\mathrm{A}}^{2025}-{\mathrm{A}}^{2020}=\left[\begin{array}{lll}0& 0& 0\\ 5& 0& 0\\ 0& 0& 0\end{array}\right]={\mathrm{A}}^6-\mathrm{A}\end{array} \end{gathered}$$