Let A=pmatrix1&0&00&1&11&0&0pmatrix.

Question

Let $A=\begin{pmatrix}1&0&0\0&1&1\1&0&0\end{pmatrix}$. Then $A^{2025}-A^{2020}$ is equal to

Accepted Answer

By induction, $A^n=\begin{pmatrix}1&0&0\n-1&1&1\1&0&0\end{pmatrix}$ for $n\ge1$.

$$A^{2025}-A^{2020}=\begin{pmatrix}1&0&0\2024&1&1\1&0&0\end{pmatrix}-\begin{pmatrix}1&0&0\2019&1&1\1&0&0\end{pmatrix}=\begin{pmatrix}0&0&0\5&0&0\0&0&0\end{pmatrix}.$$

Also $A^6-A=\begin{pmatrix}1&0&0\5&1&1\1&0&0\end{pmatrix}-\begin{pmatrix}1&0&0\0&1&1\1&0&0\end{pmatrix}=\begin{pmatrix}0&0&0\5&0&0\0&0&0\end{pmatrix}$, matching.

So $A^{2025}-A^{2020}=A^6-A$.

Solutions · 1