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We know that for any square matrix $B$ and $\epsilon$ small, we have the expansion $$ \det (I+\epsilon B) = 1+\epsilon Tr(B) + O(\epsilon^2) $$ where $Tr(B)$ is the trace of $B$, this can be generalized for a nonsingular matrix $A$ to give the expansion $$ \det (A+\epsilon B) = \det(A)\det(I+\epsilon A^{-1}B)= \det(A)(1+\epsilon Tr(A^{-1}B)+O(\epsilon^2)). $$

So my question is: are there any similar formulas when $A$ is a singular matrix? Looking at some examples, I suspect the order of the higher order term is related to the dimension of the kernel of $A$, for example, if $A= \begin{pmatrix}0&0\\0&1\end{pmatrix}$, which has a $1$ dimensional kernel, then assume $B = \begin{pmatrix}a&b\\c&d\end{pmatrix}$, we get $\det(A+\epsilon B)= \epsilon a+O(\epsilon^2)$, whereas if $A$ is the (2 by 2) zero matrix, then $\det (A+\epsilon B)$ is clearly $O(\epsilon^2)$.

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    One case which you can analyze explicitly is when $B = I$. Then $\det(A + \varepsilon B)$ is just the characteristic polynomial of $-A$. If $A$ is diagonalizable then indeed the leading term will be $\varepsilon^d$ where $d$ is the dimension of the kernel and the coefficient is $(-1)^{n-d} \operatorname{tr}(\Lambda^{n-d}(-A))$. If $A$ is not diagonalizable, the leading term might be of higher degree than the dimension of the kernel of $A$ (think about the case $A$ is a Jordan block). If $B \neq I$ but is invertible, you can multiply by $B^{-1}$ and reduce to the previous case. – 2017-02-04
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    If $B$ is not invertible, a relevant keyword to search for is "matrix pencil" since $\det(A + \varepsilon B)$ is the characteristic polynomial of a matrix pencil. – 2017-02-04
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    @levap thanks for the comments! originally I wanted to have a result like the (order of $\epsilon$) is $\ge dim \ker A$, it now looked really promising, i'll search for matrix pencil carefully... – 2017-02-04

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From Jacobi's formula

$ \left. \frac{d \det (A + \epsilon B)}{d \epsilon} \right|_{\epsilon=0} = \rm{tr}\left[ \rm{adj}(A)\, B \right] $

where $\rm{adj}(A)$ is the adjugate of $A$. Thus

$ \det (A + \epsilon B) = \det(A) + \epsilon \, \rm{tr}\left[ \rm{adj}(A)\, B \right] + O(\epsilon^2), $

which is well defined for all $A$, in particular when $A$ is singular.

If $A = \rm I$ the expression above reduces to $\det ({\rm I} + \epsilon B) = 1 + \epsilon \, {\rm tr}(B) + O(\epsilon^2)$.