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

The Correct Option is 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.