Question

Let $n\ge 2$ be an integer,
$A=\left(\begin{array}{ccc}\cos \left(2\pi /n\right)& \sin \left(2\pi /n\right)& 0\\ -\sin \left(2\pi /n\right)& \cos \left(2\pi /n\right)& 0\\ 0& 0& 1\end{array}\right)$
and $I$ is the identity matrix of order $3$. Then

• A. $A^n=I$ and $A^{n-1}\ne I$
• B. $A^m\ne I$ for any positive integer $m$
• C. $A$ is not invertible
• D. $A^m=0$ for a positive integer $m$

Answer 1

Answer: (A)

$A=\left[\begin{array}{ccc}\cos \left(\frac{2\pi }{n}\right)& \sin \left(\frac{2\pi }{n}\right)& 0\\ -\sin \left(\frac{2\pi }{n}\right)& \cos \left(\frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$
Now, $A\times A=\left[\begin{array}{ccc}\cos \left(\frac{2\pi }{n}\right)& \sin \left(\frac{2\pi }{n}\right)& 0\\ -\sin \left(\frac{2\pi }{n}\right)& \cos \left(\frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]\times$
$\left[\begin{array}{ccc}\cos \left(\frac{2\pi }{n}\right)& \sin \left(\frac{2\pi }{n}\right)& 0\\ -\sin \left(\frac{2\pi }{n}\right)& \cos \left(\frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}\cos \left(2\times \frac{2\pi }{n}\right)& \sin \left(2\times \frac{2\pi }{n}\right)& 0\\ -\sin \left(2\times \frac{2\pi }{n}\right)& \cos \left(2\times \frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}\cos \left(2\times \frac{2\pi }{n}\right)& \sin \left(2\times \frac{2\pi }{n}\right)& 0\\ -\sin \left(2\times \frac{2\pi }{n}\right)& \cos \left(2\times \frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$
Similarly
$A^n=\left[\begin{array}{ccc}\cos \left(2^{n-1}\times \frac{2\pi }{n}\right)& \sin \left(2^{n-1}\times \frac{2\pi }{n}\right)& 0\\ -\sin \left(2^{n-1}\times \frac{2\pi }{n}\right)& \cos \left(2^{n-1}\times \frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$
$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I$
and
$A^{n-1}=\left[\begin{array}{ccc}\cos \left(2^{n-2}\times \frac{2\pi }{n}\right)& \sin \left(2^{n-2}\times \frac{2\pi }{n}\right)& 0\\ -\sin \left(2^{n-2}\times \frac{2\pi }{n}\right)& \cos \left(2^{n-2}\times \frac{2\pi }{n}\right)& 0\\ 0& 0& 1\end{array}\right]$
$\ne I$
$\therefore A^m\ne I$ for any positive integer $m$.