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

The correct option is A $A^n+I$ and $A^{n-1}\ne I$

$n\ge 2$

$A=\left[\begin{array}{ccc}\frac{\cos 2\pi }{n}& \frac{\sin 2\pi }{n}& 0\\ \frac{\sin 2\pi }{n}& \frac{\cos 2\pi }{n}& 0\\ 0& 0& 1\end{array}\right]$

Let $n=2$ Then $A=\left[\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right]$

$A^2=\left[\begin{array}{ccc}1& \frac{\sin 4\pi }{2}& 0\\ \frac{\sin 4\pi }{2}& 1& 0\\ 0& 0& 1\end{array}\right]$

$A^2=I$

Verifying option A

$A^n+I$ and $A^{n-1}\ne I$

$n=2$

$A^2+I$ and $A^1\ne I$

$I+I=2I$

Hence option A is correct

Option B $A^m\ne I$ for all positive integers of m

But $A^2=I$

Option B is incorrect