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