Ans: (c)

Soloution:

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right] \\ {\mathrm{A}}^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right] \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right] = \left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 0& 0\end{array}\right] \\ {\mathrm{A}}^3=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 0& 0\end{array}\right] \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right] \\ {\mathrm{A}}^3= \left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 1\\ 1& 0& 0\end{array}\right] \\ \mathrm{Then} \mathrm{similarly} {\mathrm{A}}^{2025}=\left[\begin{array}{ccc}1& 0& 0\\ 2024& 1& 1\\ 1& 0& 0\end{array}\right] \mathrm{and} {\mathrm{A}}^{2020}=\left[\begin{array}{ccc}1& 0& 0\\ 2019& 1& 1\\ 1& 0& 0\end{array}\right] \\ \therefore {\mathrm{A}}^{2025}-{\mathrm{A}}^{2020}=\left[\begin{array}{ccc}1& 0& 0\\ 2024& 1& 1\\ 1& 0& 0\end{array}\right] - \left[\begin{array}{ccc}1& 0& 0\\ 2019& 1& 1\\ 1& 0& 0\end{array}\right] \\ =\left[\begin{array}{ccc}0& 0& 0\\ 5& 0& 0\\ 0& 0& 0\end{array}\right]={\mathrm{A}}^6-\mathrm{A} \end{gathered}$$