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 $n=2$
$P=P-[P_{ij}]_{2 \times 2} =\begin {bmatrix}p_{11} & p_{12} \\p_{21} & p_{22} \end {bmatrix}= \begin {bmatrix}\omega^{2} & \omega^{3} \\\omega^{3} & \omega^{4} \end {bmatrix} =\begin {bmatrix}\omega^{2} & 1 \\1 & \omega \end {bmatrix}$
$p^2=\begin {bmatrix}\omega_{2} & 1 \\1 & \omega \end {bmatrix} \begin {bmatrix}\omega^{2} & 1 \\1 & \omega \end {bmatrix} \Rightarrow \ $
$p^2 \begin {bmatrix}\omega^{4}+1 & \omega^{2}+\omega \\\omega^{2}+\omega & 1+\omega^{2} \end {bmatrix} \ne 0$
When $n = 3$
$p=[p_{ij}]_{3 \times 3}=$ $ \begin {bmatrix}\omega^2 & \omega^3 & \omega^4 \\\omega^3 & \omega^4 & \omega^5 \\\omega^4 & \omega^5 & \omega^6 \end {bmatrix} = \begin {bmatrix}\omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\\omega & \omega^2 & 1 \end {bmatrix}$
$p^2= \begin {bmatrix}\omega^2 & 1 & \omega \\1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \end {bmatrix} \begin {bmatrix}\omega^2 & 1 & \omega \\1 & \omega & \omega^2 \\\omega & \omega^2 & 1 \end {bmatrix} = \begin {bmatrix}0 & 0 & 0\\0 & 0 \ & 0\\0 & 0 & 0 & \end {bmatrix} = 0$
$\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.