If $\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\begin{pmatrix}d^*&-b^*\\-c^*&a^*\end{pmatrix}$ then

Question

I want to prove that for a $2\times 2$ matrix $ \begin{pmatrix} a & b \\ c& d \end{pmatrix} $ satisfying \begin{align}\tag{1} \begin{pmatrix} a & b \\ c& d \end{pmatrix}^{-1} = \begin{pmatrix} d^* & -b^* \\ -c^* & a^* \end{pmatrix} \end{align} then \begin{align} \begin{pmatrix} a & b \\ c& d \end{pmatrix} = e^{i\phi} \begin{pmatrix} A & B \\ C& D \end{pmatrix} \end{align} where $\phi,A,B,C,D$ are real and $AD-BC=1$.

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Starting from \begin{align} \begin{pmatrix} a & b \\ c& d \end{pmatrix}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ - c& a \end{pmatrix} \end{align} we have \begin{align} a^* = \frac{a}{ad-bc} \qquad b^* = \frac{b}{ad-bc} \qquad c^* = \frac{c}{ad-bc} \qquad d^* = \frac{d}{ad-bc} \end{align} but now I got confused.

Apparently I can see that for $a=e^{i\phi}A$, the above relation holds. But I don't know how it derives from $a^*=\frac a{ad-bc}$.

Answer 1

\begin{align*} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \bar{d} & -\bar{b} \\ -\bar{c} & \bar{a} \end{pmatrix} \\ &= \begin{pmatrix} a\bar{d}-b\bar{c} & b\bar{a}-a\bar{b} \\ c\bar{d}-d\bar{c} & d\bar{a}-c\bar{b} \end{pmatrix} \end{align*}

Equating "$0$":

$$ \left \{ \begin{align*} \frac{a}{\bar{a}} &= \frac{b}{\bar{b}} \\ \frac{c}{\bar{c}} &= \frac{d}{\bar{d}} \\ \end{align*} \right.$$

Let $ \left \{ \begin{align*} a &= Ae^{i\theta} \\ b &= Be^{i\theta} \end{align*} \right.$ and $ \left \{ \begin{align*} c &= Ce^{i\phi} \\ d &= De^{i\phi}\ \end{align*} \right.$ where $A,B,C,D\in \mathbb{R}$

Equating "$1$":

$$ \left \{ \begin{align*} (AD-BC)e^{i(\theta-\phi)} &= 1 \\ (AD-BC)e^{i(\phi-\theta)} &= 1 \\ \end{align*} \right. $$

$$\implies \left \{ \begin{align*} AD-BC &= 1 \\ \theta &= \phi \\ \end{align*} \right. $$