See @arnold1992ordinary section 16.3. The determinant and the trace of a matrix has the following relation.
Let $A: \mathbb R^n \to \mathbb R^n$ be a linear operator, and let $\epsilon \in \mathbb{R}.$ Then, as $\epsilon \to 0$,
$$ \det{(I+\epsilon A)}=1+\epsilon \mbox{tr}(A)+O(\epsilon^2). $$Also, $\det{e^A}=e^{\mbox{tr}(A)}$.
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Author of the notes: Antonio J. Pan-Collantes
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