Question

Let $A=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$ then find $A^{4n+1}$ for every $n\in N$.

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

Answer 1

The correct option is A $\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$

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

Now, $A^{4n+1}$

$=\left[\begin{array}{ccc}{i}^{4n+1}& 0& 0\\ 0& i^{4n+1}& 0\\ 0& 0& i^{4n+1}\end{array}\right]$ [Using property of diagonal matrix]

$=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$ [Since $i^2=-1\Rightarrow i^4=1\Rightarrow i^{4n+1}=(i^4{)}^n.i=i$]