Matrices and Determinants 1 Question 13
13. Let $\omega$ be a complex cube root of unity with $\omega \neq 0$ and $P=\left[p_{i j}\right]$ be an $n \times n$ matrix with $p_{i j}=\omega^{i+j}$. Then, $P^{2} \neq 0$ when $n$ is equal to
(2013 Adv.)
(a) 57
(b) 55
(c) 58
(d) 56
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Answer:
Correct Answer: 13. (b,c,d)
Solution:
- Here, $P =\left[p_{i j}\right]_{n \times n}$
with
$p_{i j}=w^{i+j}$
$\therefore$ When $n=1$
$ \begin{array}{rlrl} P & =\left[p_{i j}\right]_{1 \times 1}=\left[\omega^{2}\right] \\ \Rightarrow P^{2} & =\left[\omega^{4}\right] \neq 0 \end{array} $
$\therefore$ When $n=2$
$P=$ $\left[p_{i j}\right]_{2 \times 2}$
=$\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 \quad P^{2}=$ $\begin{bmatrix} \omega^{4}+1 & \omega^{2}+\omega \\ \omega^{2}+\omega & 1+\omega^{2} \end{bmatrix}$ $\neq 0$
\omega^{2}
\omega
When $n=3$
$P=$ $\left[p_{i j}\right]_{3 \times 3}$
=$\begin{bmatrix} \omega^{2} & \omega^{3} & \omega^{4} \\ \omega^{3} & \omega^{4} & \omega^{5} \\ \omega^{4} & \omega^{5} & \omega^{6} \end{bmatrix}$ =$\begin{bmatrix} \omega^{2} & 1 & \omega \\ 1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \end{bmatrix}$
$p^2$=$\begin{bmatrix} \omega^{2} & 1 & \omega \\ 1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \end{bmatrix}$ $\begin{bmatrix} \omega^{2} & 1 & \omega \\ 1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \end{bmatrix}$ =$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$=0
$\therefore \quad P^{2}=0$, when $n$ is a multiple of 3 .
$P^{2} \neq 0$, when $n$ is not a multiple of 3 .
$\Rightarrow \quad n=57$ is not possible.
$\therefore \quad n=55,58,56$ is possible.
