Determinant near identity lemma

Lemma Let $A$ be a $n\times n$ square matrix, then

$$\det (I+ϵA)=1+ϵTr(A)+o({ϵ}^2)$$

Proof Let $M=I+ϵA$.

$$\det M=\sum _{\sigma }(-1{)}^{\sigma }\prod _{i=1}^n{M}_{i{\sigma }_i}$$

Where $\sigma$ are the permutations of $[n]$. If $\sigma$ is the identity permutation, we get

$$\prod _{i=1}^n{M}_{ii}=\prod _{i=1}^n(1+ϵA_{ii})=1+ϵTr(A)+o({ϵ}^2)$$

for al the remaining permutation, at least one term of the product will be just $(ϵA_{ij})$, making only $o({ϵ}^2)$ contributions. $\square$