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