Question

For two unimodular complex numbers ${{z}_{1}}$ and ${{z}_{2}}$, the value of
${{\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}$ is equal to?
A. $\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ {{{\bar{z}}}_{1}} & {{{\bar{z}}}_{2}} \\\ \end{matrix} \right)$
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. None of these

Answer 1

Hint: In this question, the product of inverses of two matrices are needed to be found out. For that we can use the formula for the inverse of the product of two matrices and then do suitable matrix multiplication to obtain the answer to this question.

Complete step-by-step answer:
In this question, as the product of the inverses of two matrices are given, we need to use the formula that for two matrices A and B, the inverse of their product is given by
${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}.........................(1.1)$
In this case taking $A=\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)$ and $B=\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)$ in equation (1.1), we get
${{\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}={{\left( \left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right) \right)}^{-1}}...................(1.2)$
Now, the matrix product between two general matrices is given by
$\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} \\\ {{a}_{21}} & {{a}_{22}} \\\ \end{matrix} \right)\left( \begin{matrix} {{b}_{11}} & {{b}_{12}} \\\ {{b}_{21}} & {{b}_{22}} \\\ \end{matrix} \right)=\left( \begin{matrix} {{a}_{11}}{{b}_{11}}+{{a}_{12}}{{b}_{21}} & {{a}_{11}}{{b}_{12}}+{{a}_{12}}{{b}_{22}} \\\ {{a}_{21}}{{b}_{11}}+{{a}_{22}}{{b}_{21}} & {{a}_{21}}{{b}_{12}}+{{a}_{22}}{{b}_{22}} \\\ \end{matrix} \right)..................(1.3)$
So, evaluating (1.2) using the formula given in equation (1.3), we obtain

{{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}={{\left( \left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right) \right)}^{-1}}={{\left( \begin{matrix} {{z}_{1}}{{{\bar{z}}}_{1}}+{{z}_{2}}{{{\bar{z}}}_{2}} & -{{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{1}} \\\ -{{{\bar{z}}}_{2}}{{{\bar{z}}}_{1}}+{{{\bar{z}}}_{1}}{{{\bar{z}}}_{2}} & -{{{\bar{z}}}_{2}}{{z}_{2}}+{{{\bar{z}}}_{1}}{{z}_{1}} \\\ \end{matrix} \right)}^{-1}}...................(1.4)$$ Now, as $${{z}_{1}}$$ and $${{z}_{2}}$$ are given to be unimodular, we have $${{z}_{1}}{{\bar{z}}_{1}}={{\bar{z}}_{1}}{{z}_{1}}=1$$ and $${{z}_{2}}{{\bar{z}}_{2}}={{\bar{z}}_{2}}{{z}_{2}}=1$$. Using this in equation (1.4), we obtain $${{\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}={{\left( \begin{matrix} {{z}_{1}}{{{\bar{z}}}_{1}}+{{z}_{2}}{{{\bar{z}}}_{2}} & -{{z}_{1}}{{z}_{2}}+{{z}_{2}}{{z}_{1}} \\\ -{{{\bar{z}}}_{2}}{{{\bar{z}}}_{1}}+{{{\bar{z}}}_{1}}{{{\bar{z}}}_{2}} & -{{{\bar{z}}}_{2}}{{z}_{2}}+{{{\bar{z}}}_{1}}{{z}_{1}} \\\ \end{matrix} \right)}^{-1}}={{\left( \begin{matrix} 1+1 & 0 \\\ 0 & 1+1 \\\ \end{matrix} \right)}^{-1}}={{\left( \begin{matrix} 2 & 0 \\\ 0 & 2 \\\ \end{matrix} \right)}^{-1}}..............(1.5)$$ Now, the inverse of a general $2\times 2$ matrix is given by ${{\left( \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right)}^{-1}}=\dfrac{1}{ad-bc}\left( \begin{matrix} d & -b \\\ -c & a \\\ \end{matrix} \right)..................(1.6)$ Therefore, by using the formula given in equation (1.6) in equation (1.5), we obtain $$\begin{aligned} & {{\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}={{\left( \begin{matrix} 2 & 0 \\\ 0 & 2 \\\ \end{matrix} \right)}^{-1}}=\dfrac{1}{2\times 2-0\times 0}\left( \begin{matrix} 2 & -0 \\\ -0 & 2 \\\ \end{matrix} \right)=\dfrac{1}{4}\left( \begin{matrix} 2 & 0 \\\ 0 & 2 \\\ \end{matrix} \right) \\\ & =\left( \begin{matrix} \dfrac{2}{4} & 0 \\\ 0 & \dfrac{2}{4} \\\ \end{matrix} \right)=\left( \begin{matrix} \dfrac{1}{2} & 0 \\\ 0 & \dfrac{1}{2} \\\ \end{matrix} \right) \\\ \end{aligned}$$ As when a number is multiplied to a matrix, it gets multiplied in all its components. Therefore, we obtain the answer to the following question as $${{\left( \begin{matrix} {{{\bar{z}}}_{1}} & -{{z}_{2}} \\\ {{{\bar{z}}}_{2}} & {{z}_{1}} \\\ \end{matrix} \right)}^{-1}}{{\left( \begin{matrix} {{z}_{1}} & {{z}_{2}} \\\ -{{{\bar{z}}}_{2}} & {{{\bar{z}}}_{1}} \\\ \end{matrix} \right)}^{-1}}=\left( \begin{matrix} \dfrac{1}{2} & 0 \\\ 0 & \dfrac{1}{2} \\\ \end{matrix} \right)$$ Which matches option (c). Hence, option (c) is the correct answer. Note: We should be careful that we have to take the overall inverse in the last term of equation (1.4) as just multiplication of the matrices in reverse order does not give the inverse of their product. We have to take the inverse of the multiplication of the matrices in reverse order to find the inverse of their product as shown in equation (1.1).