Question

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

• A. 57
• B. 55
• C. 58
• D. 56

Answer 1

The correct option is A

57

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