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# 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 toA. 1B. 2C. -1D. -2

Last updated date: 11th Aug 2024
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Answer
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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.