$\det(I+A) = 1 + tr(A) + \det(A)$ 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 $n>3$. 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 $n>2$? 
 A: $$
\left|
  \begin{array}{ccc}
    1+a_{11} & a_{12} & a_{13} \\
    a_{21} & 1+a_{22} & a_{23} \\
    a_{31} & a_{32} & 1+a_{33} \\
  \end{array}
\right|
=
\left|
  \begin{array}{ccc}
    1 & a_{12} & a_{13} \\
    0 & 1+a_{22} & a_{23} \\
    0 & a_{32} & 1+a_{33} \\
  \end{array}
\right|
+
\left|
  \begin{array}{ccc}
    a_{11} & a_{12} & a_{13} \\
    a_{21} & 1+a_{22} & a_{23} \\
    a_{31} & a_{32} & 1+a_{33} \\
  \end{array}
\right|=
$$
$$
=
\left|
  \begin{array}{cc}
    1+a_{22} & a_{23} \\
    a_{32} & 1+a_{33} \\
  \end{array}
\right|
+
\left|
  \begin{array}{ccc}
    a_{11} & 0 & a_{13} \\
    a_{21} & 1  & a_{23} \\
    a_{31} & 0 & 1+a_{33} \\
  \end{array}
\right|
+
\left|
  \begin{array}{ccc}
    a_{11} & a_{12} & a_{13} \\
    a_{21} & a_{22} & a_{23} \\
    a_{31} & a_{32} & 1+a_{33} \\
  \end{array}
\right|
=\ldots
$$
$$
=
1+{\rm tr}A + M_{11} + M_{22} + M_{33}+\det A
$$
and further by induction.
A: 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


*

*$\det(I+tA)=(1+ta_1)(1+ta_2)\cdots(1+ta_n)= 1 + t\operatorname{tr}A+ \ldots + t^n\det A$.

A: One can formally expand $\det(I+tA)$ as a power series of $t$ and get:
$$\begin{align}
  & \det(I + tA)\\
= & \exp\left(\text{Tr}\log(I+tA)\right)\\
= & \exp\left(t\,\text{Tr}A - \frac{t^2}{2}\text{Tr}A^2 + \frac{t^3}{3}\text{Tr}A^3 + \cdots
\right)\\
= & 1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right)
+ \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right)
+ \cdots
\end{align}$$
When $A$ is a $n \times n$ matrix, the above expansion terminate at the $t^n$ term with coefficient equal to $\det A$. With this, you can obtain formula similar to what you have for $n = 2$:
$$
\det(I+tA) = \begin{cases}
1 + t\,\text{Tr}A + t^2 \det(A) & n = 2\\
\\
1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right) + t^3 \det(A) & n = 3\\
\\
1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right)
\\\;\;\;+ \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right)
+ t^4 \det(A) & n = 4\\
\end{cases}
$$
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.
