Question

Let $z_1$ and $z_2$ be two complex numbers such that $|z_1|=|z_2|=1.$ If $C={\left[\begin{array}{cc}{\overline{z}}_1& -z_2\\ {\overline{z}}_2& z_1\end{array}\right]}^{-1}{\left[\begin{array}{cc}{z}_1& z_2\\ -{\overline{z}}_2& {\overline{z}}_1\end{array}\right]}^{-1},$ then the sum of principal diagonal entries of $C$ is

• A. $1$
• B. $4$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is A $1$

Let $A=\left[\begin{array}{cc}{\overline{z}}_1& -z_2\\ {\overline{z}}_2& z_1\end{array}\right]$ and $B=\left[\begin{array}{cc}{z}_1& z_2\\ -{\overline{z}}_2& {\overline{z}}_1\end{array}\right]$

$C=A^{-1}{B}^{-1}=(BA{)}^{-1}$
$={\left(\left[\begin{array}{cc}{z}_1& z_2\\ -{\overline{z}}_2& {\overline{z}}_1\end{array}\right]\left[\begin{array}{cc}{\overline{z}}_1& -z_2\\ {\overline{z}}_2& z_1\end{array}\right]\right)}^{-1}$

$={\left[\begin{array}{cc}|z_1{|}^2+|z_2{|}^2& 0\\ 0& |z_1{|}^2+|z_2{|}^2\end{array}\right]}^{-1}$

$={\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]}^{-1}$

$=\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right]$
$\therefore$ Sum of diagonal elements $=1$