Solution
= Solution
$A=\begin{pmatrix}x_3&x_1-ix_2\\x_1+ix_2&-x_3\end{pmatrix}$, so $\operatorname{tr}A=0$ and $\det A=-|x|^2$. Unitary conjugation preserves Hermiticity, trace, and <determinant>; every traceless Hermitian $2\times2$ <matrix> is $x\prime\cdot\sigma$ with real $x\prime$, and $|x\prime|=|x|$.