1

There are 1 best solutions below

1
On

It is because $\newcommand{\eps}{\varepsilon}\newcommand{\tr}{\operatorname{tr}}$ $$ \det(A+\eps B)=\det(A)\det(I+\eps A^{-1}B)=\det(A)(1+\eps \tr(A^{-1}B)+O(\eps^2)) $$ The derivative of the determinant follows, then insert the ODE.