Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 66 / 4 / b / Solution

Set $S=A+B$ and $E(t)=F(t)-e^{tS}$. Since $F(0)=I$, $E(0)=0$. Differentiate the ordered exponential products, using that each matrix commutes with its own exponential:Here the commutator convention is $[X,Y]=XY-YX$; this fixes both signs. The error satisfies $E^{\prime }=SE+D$, so the variation-of-constants formula givesThis is the symmetrized exponential-splitting defect identity. Symmetry of $A,B$ was not needed for the identity itself; it will be used to bound their exponentials in part (c).