Question

For two unimodular complex numbers ${{z}_{1}}\text{ and }{{z}_{2}}$, ${{\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]}^{-1}}{{\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]}^{-1}}$ is
\[\begin{align}
  & A.\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{{{z}_{1}}} & \overline{{{z}_{2}}} \\
\end{matrix} \right] \\
 & B.\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] \\
 & C.\left[ \begin{matrix}
   \dfrac{1}{2} & 0 \\
   0 & \dfrac{1}{2} \\
\end{matrix} \right] \\
 & D.\text{ None of these} \\
\end{align}\]

Answer 1

Hint: In this question, we are given two unimodular complex numbers ${{z}_{1}}\text{ and }{{z}_{2}}$ and we have to find value of given ${{\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]}^{-1}}{{\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]}^{-1}}$. We will first suppose these two matrix as variable and find their inverse using ${{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA$ where $\left| A \right|$ is determinant of A and adjA is adjoint of a. Adjoint of A will be found by taking the transpose of the cofactor matrix. We will use property as $z\overline{z}={{\left| z \right|}^{2}}$ where $\overline{z}$ is conjugate of z. Also cofactor matrix of $2\times 2$ matrix $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ is given by $\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]$.

Complete step by step answer:
Here, we have to find value of ${{\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]}^{-1}}{{\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]}^{-1}}$.
Since, it is given that, complex numbers ${{z}_{1}}\text{ and }{{z}_{2}}$ are unimodular, therefore, their modulus is 1, that is $\left| {{z}_{1}} \right|\text{=1 and }\left| {{z}_{2}} \right|=1$.
Now, let us suppose
\[A=\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]\text{ and }B=\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]\]
Hence, we have to find ${{A}^{-1}}{{B}^{-1}}$. For finding ${{A}^{-1}}$. Let's first find $\left| A \right|$.
\[\begin{align}
  & \left| A \right|=\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right] \\
 & \Rightarrow \overline{{{z}_{1}}}{{z}_{1}}-\left( -{{z}_{2}}\overline{{{z}_{2}}} \right) \\
 & \Rightarrow \overline{{{z}_{1}}}{{z}_{1}}+\overline{{{z}_{2}}}{{z}_{2}} \\
\end{align}\]
Since $z\overline{z}={{\left| z \right|}^{2}}$,
Therefore, $\left| A \right|={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}$.
Now, we know that $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1$.
So, $\left| A \right|={{\left| 1 \right|}^{2}}+{{\left| 1 \right|}^{2}}=1+1=2$.
Now, let's find adjA.
Adjoint of any matrix is the transpose of the cofactor matrix.
As we know, cofactor matrix of $2\times 2$ matrix $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ is given by $\left[ \begin{matrix}
   d & -c \\
   -b & a \\
\end{matrix} \right]$.
Therefore, $\text{cofactor matrix of }A=\left[ \begin{matrix}
   {{z}_{1}} & \overline{-{{z}_{2}}} \\
   -{{z}_{2}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]$
\[\text{Adjoint of }A={{\left[ \begin{matrix}
   {{z}_{1}} & \overline{-{{z}_{2}}} \\
   -{{z}_{2}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]\]
Hence, \[{{A}^{-1}}=\dfrac{1}{\left| A \right|}adjA=\dfrac{1}{2}\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]\cdots \cdots \cdots \cdots \left( 1 \right)\]
For finding ${{B}^{-1}}$. Let us first find $\left| B \right|$.
\[\begin{align}
  & \left| B \right|=\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right] \\
 & \Rightarrow {{z}_{1}}\overline{{{z}_{1}}}-\left( -{{z}_{2}}\overline{{{z}_{2}}} \right) \\
 & \Rightarrow {{z}_{1}}\overline{{{z}_{1}}}+{{z}_{2}}\overline{{{z}_{2}}} \\
\end{align}\]
Since $z\overline{z}={{\left| z \right|}^{2}}$ therefore, $\left| B \right|={{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}}$.
Now, we know that $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1$ so, $\left| B \right|={{\left| 1 \right|}^{2}}+{{\left| 1 \right|}^{2}}=1+1=2$.
Let's find adjB.
$\text{cofactor matrix of }B=\left[ \begin{matrix}
   \overline{{{z}_{1}}} & \overline{{{z}_{2}}} \\
   {{z}_{2}} & {{z}_{1}} \\
\end{matrix} \right]$
\[\text{Adjoint of }B={{\left[ \begin{matrix}
   \overline{{{z}_{1}}} & \overline{{{z}_{2}}} \\
   {{z}_{2}} & {{z}_{1}} \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   \overline{{{z}_{1}}} & {{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]\]
Hence, \[{{B}^{-1}}=\dfrac{1}{\left| B \right|}adjB=\dfrac{1}{2}\left[ \begin{matrix}
   \overline{{{z}_{1}}} & {{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]\cdots \cdots \cdots \cdots \left( 2 \right)\]
We want ${{A}^{-1}}{{B}^{-1}}$ therefore, let's multiply (1) and (2) together, we get:
\[\begin{align}
  & {{A}^{-1}}{{B}^{-1}}=\dfrac{1}{4}\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]\left[ \begin{matrix}
   \overline{{{z}_{1}}} & {{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right] \\
 & \Rightarrow \dfrac{1}{4}\left[ \begin{matrix}
   {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}} & 0 \\
   0 & {{\left| {{z}_{1}} \right|}^{2}}+{{\left| {{z}_{2}} \right|}^{2}} \\
\end{matrix} \right] \\
 & \Rightarrow \dfrac{1}{4}\left[ \begin{matrix}
   {{1}^{2}}+{{1}^{2}} & 0 \\
   0 & {{1}^{2}}+{{1}^{2}} \\
\end{matrix} \right] \\
 & \Rightarrow \dfrac{1}{4}\left[ \begin{matrix}
   2 & 0 \\
   0 & 2 \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   \dfrac{1}{2} & 0 \\
   0 & \dfrac{1}{2} \\
\end{matrix} \right] \\
\end{align}\]
Hence, \[{{\left[ \begin{matrix}
   \overline{{{z}_{1}}} & -{{z}_{2}} \\
   \overline{{{z}_{2}}} & {{z}_{1}} \\
\end{matrix} \right]}^{-1}}{{\left[ \begin{matrix}
   {{z}_{1}} & {{z}_{2}} \\
   \overline{-{{z}_{2}}} & \overline{{{z}_{1}}} \\
\end{matrix} \right]}^{-1}}=\left[ \begin{matrix}
   \dfrac{1}{2} & 0 \\
   0 & \dfrac{1}{2} \\
\end{matrix} \right]\]

So, the correct answer is “Option C”.

Note: Students should carefully calculate determinants of the matrix since small sign mistakes can completely change the answer. Students can get confused between A and B matrices, ${{z}_{1}}\text{ and }{{z}_{2}}$ their conjugate and negative signs. So take care while copying them and solving them. For finding adjoint of the matrix, don’t forget to find transpose.