Question

Let $ \omega $ be the complex cube root of the unity with and $ P=\left[ {{p}_{ij}} \right] $ be a $ 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

Hint:
We will at first form the general matrix of $ n\times n $ as each element of the matrix is given by $ {{p}_{ij}}={{\omega }^{i+j}} $ , where i and j are the row and column of that element and then we will multiply matrix P with itself and use the property that $ {{\omega }^{3}}=1 $ and $ 1+\omega +{{\omega }^{2}}=0 $ because it is given in the question that $ \omega $ be the complex cube root of the unity.

Complete step by step answer:
We know from the question that each element of the matrix $ P=\left[ {{p}_{ij}} \right] $ is given by $ {{p}_{ij}}={{\omega }^{i+j}} $ , where i and j are the row and column number corresponding to that element.
So, the matrix P is given as:
\[P=\left[ \begin{matrix}
   {{\omega }^{2}} & {{\omega }^{3}} & {{\omega }^{4}} & \cdots & {{\omega }^{n+1}} \\
   {{\omega }^{3}} & {{\omega }^{4}} & {{\omega }^{5}} & \cdots & {{\omega }^{n+2}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   {{\omega }^{n}} & {{\omega }^{n+1}} & {{\omega }^{n+2}} & \cdots & {{\omega }^{2n-1}} \\
   {{\omega }^{n+1}} & {{\omega }^{n+2}} & {{\omega }^{n+3}} & \cdots & {{\omega }^{2n}} \\
\end{matrix} \right]\]
Now, we will take common $ {{\omega }^{2}},{{\omega }^{3}},{{\omega }^{4}}... $ and so on from 1 , 2, 3, …. and so, on rows respectively.
So, we will get:
 $ \Rightarrow P=\left( {{\omega }^{2}}\times {{\omega }^{3}}\times .......\times {{\omega }^{n}}\times {{\omega }^{n+1}} \right)\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
 $ \Rightarrow P={{\omega }^{2+3+4+...+n+(n+1)}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
So, after simplifying P we will get:
 $ \Rightarrow P={{\omega }^{(1+2+3+4+...+n)+n}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
Since, we know that 1+2+3+4+….+n = $ \dfrac{n\left( n+1 \right)}{2} $ . So, we will get:
 $ \Rightarrow P={{\omega }^{\dfrac{n\left( n+1 \right)}{2}+n}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
 $ \Rightarrow P={{\omega }^{\dfrac{n\left( n+3 \right)}{2}}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
Now, we will multiply P by itself to get $ {{P}^{2}} $ .
So, $ {{P}^{2}}={{\omega }^{\dfrac{n\left( n+3 \right)}{2}}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right]\times {{\omega }^{\dfrac{n\left( n+3 \right)}{2}}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
 $ \Rightarrow {{P}^{2}}={{\omega }^{n\left( n+3 \right)}}\left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right]\times \left[ \begin{matrix}
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   \cdots & \cdots & \cdots & \cdots & \cdots \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
   1 & \omega & {{\omega }^{2}} & \cdots & {{\omega }^{n-1}} \\
\end{matrix} \right] $
We know that when we multiply two matrices, then each element is obtained by term by term multiplication of entries of the ‘i’ row and the ‘j’ column of first and second matrix respectively and then add them.
Since, $ {{P}^{2}}\ne 0 $
So, we have $ \left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+...+{{\omega }^{n-1}} \right) $ common in all the entries of $ {{P}^{2}} $ .
So, $ \left( 1+\omega +{{\omega }^{2}}+{{\omega }^{3}}+...+{{\omega }^{n-1}} \right)\ne 0 $
 $ \Rightarrow \left( \dfrac{1-{{\omega }^{n}}}{1-\omega } \right)\ne 0 $
Since, it is given in the question that $ \omega $ be the complex cube root of the unity, so $ {{\omega }^{3}}=1 $ .
So, we can say that when n is multiple of 3 then $ {{P}^{2}} $ will become equal to zero.
Now, after analyzing the option we can see that 57 is divisible by 3. So, except 57 i.e. option (a) all other options are true.
Hence, the correct answer is an option (b), (c) and (d).

Note:
 We are required to note that when we multiply a matrix by any constant then it gets multiplied to a particular row or column, not to the whole matrix and it is similarly applicable when we take out something common from a matrix.