Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 31A / b / Solution

The $k^{-1}$ coefficient of the first equation is ${\mu }_{1x}+i[\sigma ,{\mu }_2]=Q{\mu }_1$. Substitution in the expression for $B$ gives $B=-2{\mu }_{1x}$. Decompose ${\mu }_1$ into diagonal and off-diagonal parts. Since $Q=i[\sigma ,{\mu }_1]$, its off-diagonal part is $iQ\sigma /2$. ThusThe commutator in the $k^{-1}$ equation has zero diagonal; multiplication of two off-diagonal two-by-two matrices is diagonal. Combining the two parts proves