$\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$? Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily  verifies that
\begin{equation}
\det(I+A) = 1 + tr(A) + \det(A)
\end{equation}
or
\begin{equation}
\det(I+tA) = 1 +  t\  tr(A) + t^2\det(A)
\end{equation}
for some scalar $t \in \mathbb{R}$.

I have tried to see if there exists a similar formula for $n&gt;3$. This is a natural question. But the calculations are very big and difficulty to see. Then I do the answer. Is there a similar formula for $n&gt;2$?
If $a_1,a_2,\ldots,a_n$ are the eigenvalues of $A$ then the characteristic polynomial of $A$ is 
$$
\det(tI-A)=(t-a_1)(t-a_2)\cdots(t-a_n).$$ 
Therefore


$\det(I+A)=(1+a_1)(1+a_2)\cdots(1+a_n) = 1 + \operatorname{tr}A+ \ldots + \det A$


and


$\det(I+tA)=(1+ta_1)(1+ta_2)\cdots(1+ta_n)= 1 + t\operatorname{tr}A+ \ldots + t^n\det A$.
it is like the development of the characteristic polynomial (each term of the polynomial is a combination of symmetric sub-determinants of the initial matrix), I think you can easily find a proof of this fact on the internet. 
Or you can even prove it by yourself, using the multi-linearity of the determinant, and the fact that it is alternate.