Question

If $A=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$ then $A^{4n+1}=......,nϵN$

• A. $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
• B. $\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right]$
• C. $\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$
• D. $\left[\begin{array}{ccc}-i& 0& 0\\ 0& -i& 0\\ 0& 0& -i\end{array}\right]$

Answer 1

Answer: (C)

$\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$

$A=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$
take $i$ common, we get,
$A=i\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=iI$
$\therefore A^{4n+1}=(iI{)}^{4n+1}=i^{4n+1}(I{)}^{4n+1}=i^{4n}i\left(I\right)[\because I^m=1$ where $m\in N]$
$=\left(1\right)\left(i\right)\left(I\right)=iI=A[\because i^{4n}=1$ where $N\in N]$
Hence option 'C' is correct choice