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let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be $n\times n$ matrix with $$p_{ij} = \omega^{i+j}$$ Then $P^2$ is not equal to zero where $n$ is equal to

$ (a) \ 57 \\ (b)\ 55,58,56 \\ (c)\ 56,48,57 \\ (d) \ 90,56,57$

Jaideep Khare
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Anshi
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1 Answers1

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Trick: The shortest approach, in an MCQ scenario would be to work out $P^2$ for a few basic matrices.

That is, for $n=2$, $P^2$ would not be a zero matrix.

For $n=3$, the matrix $P$ according to the prescribed condition is

$$\begin{bmatrix} \omega^2 & 1 & \omega \\ 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \end{bmatrix}$$

Using basic matrix multiplication and applying the identity $1+\omega+\omega^2=0$, it is easily seen that $P^2$ is the zero matrix.

Keeping this in mind, we can infer a general rule:

for $n \ne 3m, m\in \mathbb N\implies P^2 \ne 0$

From this you can workout the options.

Ananth Kamath
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    [+1] You have found the trick (of this very tricky and un-paedagogical exercice)! Maybe, it should be said that if you call $P_3$ the $3 \times 3$ matrix that you have used, if one has a $3n \times 3n$ matrix, call it $Q$, of the general type, it can be partitionned into $n \times n$ blocks constituted by $P_3$ matrices, explaining that in this case $Q^2=0$. – Jean Marie Jun 08 '17 at 16:42
  • ... thus the good answer is (b). I am curious to know the book you are working with... – Jean Marie Jun 09 '17 at 04:54
  • But what should be the answer – Anshi Jun 09 '17 at 06:45
  • How ? Answer is b . please give me more illustration – Anshi Jun 09 '17 at 06:47
  • @Anshi Look for the multiples of $3$ in each option. Relate it with the general rule obtained. – Ananth Kamath Jun 09 '17 at 07:05
  • How 55 is possible – Anshi Jun 30 '17 at 07:29
  • @Anshi $55$ is possible for the same reason $58$ and $56$ is possible. Are these numbers multiples of $3$? What can we say if they are not? Think about it. – Ananth Kamath Jul 01 '17 at 03:48