Question

Let $A=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 1& 0& 0\end{array}\right).$ Then $A^{2025}-A^{2020}$ is equal to

• A. $A^6-A$
• B. $A^5$
• C. $A^5-A$
• D. $A^6$

Answer 1

Answer: (A)

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