I am learning (Weyl) spinor formalism from Müller-Kirsten and Wiedemann's Introduction to Supersymmetry (2nd Ed., WS, 2010, here). I am quite confused about the transformation between left-handed spinors and right-handed spinors.
On P31, left-handed spinors are defined to transform under self-representation $M\in SL(2,\mathbb{C})$: \begin{equation} \psi^{\prime}_A=M_A^B \psi_B.\tag{1.60} \end{equation}
On P32, right-handed spinors are defined to transform under complex conjugate self-representation $M^{*},M\in SL(2,\mathbb{C})$: \begin{equation} \bar{\psi}^{\prime}_\dot{A}=(M^*)_\dot{A}^\dot{B} \bar{\psi}_\dot{B} \qquad (1.62) \end{equation}
An equivalent representation to complex conjugate self-representation is discussed on P38: \begin{equation} \bar{\psi}^{\prime\dot{A}}=(M^{*-1T})^\dot{A}_\dot{B} \bar{\psi}^\dot{B} \qquad (1.81) \end{equation}
The authors claim that: \begin{equation} \psi^{A}=\bar{\psi}^{*}_{\dot{A}}\qquad (1.200) \end{equation} or equivalently:
\begin{equation} \psi^{A}=\bar{\psi}^{*}_{\dot{B}}(\bar{\sigma}^0)^{\dot{B}A}\qquad (1.199) \end{equation}
The right-hand sides in the two equations are the same since $(\bar{\sigma}^0)^{\dot{B}A}$ equals identity matrix in components.
Here is my question: it seems like (1.200) contradicts the definitions (1.60) and (1.81). To be precise, for a $M\in SL(2,\mathbb{C})$, we can write the following: \begin{align} &\psi^{\prime}=M\psi\\ &\bar{\psi^{\prime}}=(M^{*-1T})\bar{\psi}\\ &\psi^{*}=\bar{\psi} \end{align} where I regard the lower undotted index and upper dotted index as column indices to rewrite the above relations in matrix form (and leave indices implicit). We can deduce that \begin{equation} M=M^{-1T} \end{equation} which is NOT the property of matrices in $SL(2,\mathbb{C})$. Is there any way out?