Question

Let $\omega$ be a complex cube root of unity with $\omega \ne 0$ and $P=[P_{ii}]$ be an $n\times n$ matrix with $p_{ij}={\omega }^{i+j}$ Then, $P^2\ne 0$ when $n$ =

• A. 57
• B. 55
• C. 58
• D. 56

Answer 1

Answer: (B), (C), (D)

Here, $P=P-[P_{ij}{]}_{1\times 1}$ with $P_{ij}=w^{i+j}$
$\therefore$ When $n=1$
$P-[P_{ij}{]}_{n\times n}=[{\omega }^2]\Rightarrow P^2=[{\omega }^2]\ne 0$
$\therefore$ when $n=2$
$P=P-[P_{ij}{]}_{2\times 2}=\left[\begin{array}{cc}{p}_{11}& p_{12}\\ p_{21}& p_{22}\end{array}\right]=\left[\begin{array}{cc}{\omega }^2& {\omega }^3\\ {\omega }^3& {\omega }^4\end{array}\right]=\left[\begin{array}{cc}{\omega }^2& 1\\ 1& \omega \end{array}\right]$
$p^2=\left[\begin{array}{cc}{\omega }_2& 1\\ 1& \omega \end{array}\right]\left[\begin{array}{cc}{\omega }^2& 1\\ 1& \omega \end{array}\right]\Rightarrow$
$p^2\left[\begin{array}{cc}{\omega }^4+1& {\omega }^2+\omega \\ {\omega }^2+\omega & 1+{\omega }^2\end{array}\right]\ne 0$
When $n=3$
$p=[p_{ij}{]}_{3\times 3}=$ $\left[\begin{array}{ccc}{\omega }^2& {\omega }^3& {\omega }^4\\ {\omega }^3& {\omega }^4& {\omega }^5\\ {\omega }^4& {\omega }^5& {\omega }^6\end{array}\right]=\left[\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\end{array}\right]$
$p^2=\left[\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\end{array}\right]\left[\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\end{array}\right]=\left[\begin{array}{cccc}0& 0& 0& \\ 0& 0& 0& \\ 0& 0& 0& \end{array}\right]=0$
$\therefore p^2=0,$ when $n$ is a multiple of $3$.
$p^2\ne 0$, when $n$ is not a multiple of $3$.
$\Rightarrow n=57$ is not possible
$\therefore n=55,58,56$ is possible.