Matrices in SU(2)

Question

Let $M=\begin{bmatrix} a & c \\ b & d \end{bmatrix}\in SU(2)$. we have $M\overline M^t= \begin{bmatrix} aa^*+cc^* & ab^*+cd^* \\ ba^*+dc^* & bb^*+dd^* \end{bmatrix}$, and we must have $M\overline M^t=I $.

I wonder how to for these four quadratic equations to get $c=-\overline b$ and $d=\overline a$.

So far I am only able to conclude $a,d$ and $b,c $ have the same magnitudes.

Answer 1

Use that $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$$Imposing $ad-bc=1$ and setting this equal to $$\begin{pmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d}\end{pmatrix}$$gives the desired conclusion.

Answer 2

You probably forgot the condition $\det=1$. This implies that the inverse of your matrix is $$\begin{bmatrix} d & -b\\ -c & a \end{bmatrix}$$ Since that must be equal to $A^T$, you get that this is equal to \begin{bmatrix} \bar a & \bar c \\ \bar b & \bar d \end{bmatrix}

So $a=\bar d, \bar c=-b, \bar b=-c$.