Question

If $A=\left[\begin{array}{ccc}0& 0& x\\ 0& x& 0\\ x& 0& 0\end{array}\right]$, then $A^{2n+1}\left(nϵN\right)=$

• A. $\left[\begin{array}{ccc}{x}^{2n+1}& 0& 0\\ 0& x^{2n+1}& 0\\ 0& 0& x^{2n+1}\end{array}\right]$
• B. $\left[\begin{array}{ccc}0& 0& x^{2n+1}\\ 0& x^{2n+1}& 0\\ x^{2n+1}& 0& 0\end{array}\right]$
• C. $\left[\begin{array}{ccc}{x}^n& 0& 0\\ 0& x^n& 0\\ 0& 0& x^n\end{array}\right]$
• D. $\left[\begin{array}{ccc}0& 0& x^n\\ 0& x^n& 0\\ x^n& 0& 0\end{array}\right]$

Answer 1

Answer: (B)

$\left[\begin{array}{ccc}0& 0& x^{2n+1}\\ 0& x^{2n+1}& 0\\ x^{2n+1}& 0& 0\end{array}\right]$

$A^2=\left[\begin{array}{ccc}0& 0& x\\ 0& x& 0\\ x& 0& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& x\\ 0& x& 0\\ x& 0& 0\end{array}\right]=\left[\begin{array}{ccc}{x}^2& 0& 0\\ 0& x^2& 0\\ 0& 0& x^2\end{array}\right]$

and $A^3=A^2\times A=\left[\begin{array}{ccc}{x}^2& 0& 0\\ 0& x^2& 0\\ 0& 0& x^2\end{array}\right]\left[\begin{array}{ccc}0& 0& x\\ 0& x& 0\\ x& 0& 0\end{array}\right]=\left[\begin{array}{ccc}0& 0& x^3\\ 0& x^3& 0\\ x^3& 0& 0\end{array}\right]$

so, $A^n=\left[\begin{array}{ccc}{x}^n& 0& 0\\ 0& x^n& 0\\ 0& 0& x^n\end{array}\right]$ for $n=$ even

and for $n=$ odd, $A^n=\left[\begin{array}{ccc}0& 0& x^n\\ 0& x^n& 0\\ x^n& 0& 0\end{array}\right]$

since $2n+1\Rightarrow$ odd

so, $A^{2n+1}=\left[\begin{array}{ccc}0& 0& x^{2n+1}\\ 0& x^{2n+1}& 0\\ x^{2n+1}& 0& 0\end{array}\right]$

Hence, the answer is $\left[\begin{array}{ccc}0& 0& x^{2n+1}\\ 0& x^{2n+1}& 0\\ x^{2n+1}& 0& 0\end{array}\right].$