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

The correct option is A $A^6-A$

$|A-\lambda I|=\left|\begin{array}{ccc}1-\lambda & 0& 0\\ 0& 1-\lambda & 1\\ 1& 0& -\lambda \end{array}\right|=0$
$\Rightarrow {\lambda }^3-2{\lambda }^2+\lambda =0$

$A^3-2A^2+A=0$
$\Rightarrow A^2-A=A^3-A^2=A^4-A^3=A^5-A^4=A^6-A^5=A^7-A^6$
So, $A^7-A^2=A^6-A$
$\Rightarrow A^8-A^3=A^7-A^2=A^6-A$
And so on.
$\therefore A^{2025}-A^{2020}=A^6-A$