Question

Let $ \omega $ be the complex number $ \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} $ . Then the number of distinct complex number z satisfying $ \left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=0 $ is equal to
A. 1
B. 2
C. -1
D. -2

Answer 1

Hint:
We first apply the row-column operations to simplify the given determinant and then expand it along a column. We use the identities like $ 1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1 $ to find the cubic equation of z. we solve the equation through factorisation and find the number of distinct complex number z satisfying $ \left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=0 $ .

Complete step by step answer:
Let $ \omega $ be the complex number $ \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} $ . $ \omega $ is the complex root of unity.
We have the identities that $ 1+\omega +{{\omega }^{2}}=0;{{\omega }^{3}}=1 $ .
We have been given the equation $ \left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=0 $ . The determinant value is 0.
We first try to expand the determinant using some row-column operations.
We apply the operation $ {{C}_{1}}^{'}={{C}_{1}}+{{C}_{2}}+{{C}_{3}} $ .
So, $ \left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=\left| \begin{matrix}
   z+1+\omega +{{\omega }^{2}} & \omega & {{\omega }^{2}} \\
   z+1+\omega +{{\omega }^{2}} & z+{{\omega }^{2}} & 1 \\
   z+1+\omega +{{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right| $ .
Now we take the common term $ z+1+\omega +{{\omega }^{2}} $ out of the first column.
So, \[\left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=\left( z+1+\omega +{{\omega }^{2}} \right)\left| \begin{matrix}
   1 & \omega & {{\omega }^{2}} \\
   1 & z+{{\omega }^{2}} & 1 \\
   1 & 1 & z+\omega \\
\end{matrix} \right|\].
Now we apply the operation $ {{R}_{2}}^{'}={{R}_{2}}-{{R}_{1}};{{R}_{3}}^{'}={{R}_{3}}-{{R}_{1}} $ . We have $ 1+\omega +{{\omega }^{2}}=0 $ .
So, \[\left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=\left( z \right)\left| \begin{matrix}
   1 & \omega & {{\omega }^{2}} \\
   0 & z+{{\omega }^{2}}-\omega & 1-{{\omega }^{2}} \\
   0 & 1-\omega & z+\omega -{{\omega }^{2}} \\
\end{matrix} \right|\].
Now we expand the determinant along the first column and get
\[z\left[ \left( z+{{\omega }^{2}}-\omega \right)\left( z+\omega -{{\omega }^{2}} \right)-\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right) \right]=0\].
We simplify the equation and get
\[\begin{align}
  & \Rightarrow z\left[ {{z}^{2}}-{{\left( \omega -{{\omega }^{2}} \right)}^{2}}-1+{{\omega }^{3}}-\omega -{{\omega }^{2}} \right]=0 \\
 & \Rightarrow z\left[ {{z}^{2}}-{{\omega }^{2}}-{{\omega }^{4}}+2{{\omega }^{3}}+1 \right]=0 \\
 & \Rightarrow z\left[ {{z}^{2}}-{{\omega }^{2}}-\omega +2+1 \right]=0 \\
 & \Rightarrow z\left[ {{z}^{2}}+4 \right]=0 \\
\end{align}\]
We apply factorisation and get the values of z as $ z=0,\pm 2i $ where $ i=\sqrt{-1} $ .
Therefore, the number of distinct complex number z satisfying $ \left| \begin{matrix}
   z+1 & \omega & {{\omega }^{2}} \\
   \omega & z+{{\omega }^{2}} & 1 \\
   {{\omega }^{2}} & 1 & z+\omega \\
\end{matrix} \right|=0 $ is equal to 2. The correct option is B.

Note:
We need to always remember that in the case of determinant the multiplication and division happen only for a single row or column. In the case of the matrix, it happens for all the elements. Multiple row-column operations in a single step are not possible as the consecutive changes have to be followed.