Question

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

• 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

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

$A=\left(\begin{array}{ccc}i& 0& o\\ 0& i& 0\\ 0& 0& i\end{array}\right)$

$\therefore A^2=A.A=\left(\begin{array}{ccc}i& 0& o\\ 0& i& 0\\ 0& 0& i\end{array}\right)\left(\begin{array}{ccc}i& 0& o\\ 0& i& 0\\ 0& 0& i\end{array}\right)=\left(\begin{array}{ccc}-1& 0& o\\ 0& -1& 0\\ 0& 0& -1\end{array}\right)$

$\therefore A^4=A^2.A^2=\left(\begin{array}{ccc}-1& 0& o\\ 0& -1& 0\\ 0& 0& -1\end{array}\right)\left(\begin{array}{ccc}-1& 0& o\\ 0& -1& 0\\ 0& 0& -1\end{array}\right)=\left(\begin{array}{ccc}1& 0& o\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=I$

$⟹A^{4n+1}=A^{4n}.A=(A^4{)}^n.A=(I{)}^n.A=I.A=A$

$\therefore A^{4n+1}=A=\left(\begin{array}{ccc}i& 0& o\\ 0& i& 0\\ 0& 0& i\end{array}\right)$