Find the expansion for $\det(I+\epsilon A)$ where $\epsilon$ is small without using eigenvalue. I'm taking a linear algebra course and the professor included the problem that prove
$$
\rm{det}(I+\epsilon A) = 1 + \epsilon\,\rm{tr}\,A + o(\epsilon)
$$
Since the professor hasn't covered the concept of eigenvalue, we are not recommended to mention eigenvalue in the solution. How to prove the equation by only using Laplace theorem?
 A: We note that the determinant is defined by the Leibniz formula:
$$\det(\mathbf{I}+\epsilon\mathbf{A})=\sum_{\sigma \in S_{n}}\prod_{i=1}^{n}(\mathbf{I}+\epsilon \mathbf{A})_{i,\sigma_{i}} \cdot \operatorname{sgn}(\sigma)$$
We note that along the main diagonal we have:
$$\prod_{i=1}^{n}(\mathbf{I}+\epsilon \mathbf{A})_{i,i}=1+\epsilon (A_{11}+\cdots+A_{nn})+\mathcal{O}(\epsilon^{2})$$
And along the other possible permutations we have at best $\mathcal{O}(\epsilon^{2})$ as there must be at least 2 off-diagonal entries for it to be a vaid permutation. Therefore we can say:
$$\det(\mathbf{I}+\epsilon\mathbf{A})=1+\epsilon\operatorname{Tr}(\mathbf{A})+\mathcal{O}(\epsilon^{2})$$
As required. 
A: Consider this expression of the determinant:
$$
\det(B) = \sum_{\sigma \in S_n} (-1)^\sigma \prod_{i=1}^n b_{i,\sigma_i}
$$
For $\sigma=id$, we have the term $b_{11}b_{22}\cdots b_{nn}$. Now take $B=I+\epsilon A$ and this term becomes $(1+\epsilon a_{11})(1+\epsilon a_{22})\cdots(1+\epsilon a_{nn})=1+\epsilon \, \rm{tr}\,A+$ higher-order terms.
All other terms in the expression above contain $\epsilon^{2}$.
A: How this is done would depend on which characterizations of the determinant function you've seen.  One of those is
$$
\sum\left\{ \prod_{k=1}^n (\pm1)a_{k,\sigma(k)} : \sigma\text{ permutes }\{1,\ldots,n\} \right\}
$$
where the sign is $+1$ if $\sigma$ is an even permutation and $-1$ if it is odd.  When $\sigma$ is the identity permutation, the product is
$$
(1+\varepsilon a_{11})\cdots(1+\varepsilon a_{nn}) = 1 + \varepsilon(a_{11}+\cdots+a_{nn}) + \varepsilon^2(\cdots) + \varepsilon^3(\cdots)+\cdots.
$$
When $\sigma$ is any other permutation then the product is
$$
\varepsilon a_{k,\sigma(k)} \varepsilon a_{j,\sigma(j)}\cdots+\cdots,
$$
i.e. there are at least two indices $k$ for which $\sigma(k)\ne k$, so that $\varepsilon^2$ is a factor in this term.  Thus the determinant is
$$
1+\varepsilon\operatorname{tr}(A)+(\text{terms of degree $\ge2$ in $\varepsilon$}).
$$
