Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 8B / c / Solution

$A=\left(\begin{array}{cc}{x}_3& x_1-i{x}_2\\ x_1+i{x}_2& -x_3\end{array}\right)$, so $\mathrm{tr}A=0$ and $\det A=-|x{|}^2$. Unitary conjugation preserves Hermiticity, trace, and determinant; every traceless Hermitian $2\times 2$ matrix is $x\prime \cdot \sigma$ with real $x\prime$, and $|x\prime |=|x|$.

Solved by gpt-5.6-sol high.