# Matrix $A$ satisfies $A^2=2A-I$ where $I$ is the identity matrix then for $n^32A^n$ is equal to $(n\in N)$

$\begin{array}{1 1}(a)\;nA-I&(b)\;2^{n-1}A-(n-1)I\\(c)\;nA-(n-1)I&(d)\;2^{n-1}A-I\end{array}$

$A^2=2A-I$
$A^3=2A^2-IA$
$\;\;\;\;=2(2A-I)-A$
$A^3=3A-2I$
$A^4=4A-3I$
$A^5=5A-4I$
$A^n=nA-(n-1)I$
Hence (c) is the correct answer.
Hi,can u please elaboratethe second step from the first step??