what would the value of determinant of a matrix be if a specific entry changed? 
What will the value of determinant of matrix $A=\pmatrix{1&3&4\\5&2&a\\6&-2&3}$ be change if we change $a$ to $a+2$.

This is an easy problem because $|A|=-127+20a$ and if we did that changing, we would get $-87+20a$. My question here is:
Can we do this request by doing another way? Where does that $+40$ come from regarding the whole entries of the matrix? Thanks
 A: Expand the $|A|$ along column 3.
\begin{align}
|A|&=
\begin{vmatrix}
1&3&4\\
5&2&a\\
6&-2&3
\end{vmatrix} \\
&=
4
\begin{vmatrix}
5&2\\
6&-2
\end{vmatrix}-a
\begin{vmatrix}
1&3\\
6&-2
\end{vmatrix}+3
\begin{vmatrix}
1&3\\
5&2
\end{vmatrix} \\
&=4(-22)-a(-20)+3(-13) \\
&=-127+20a
\end{align}
A: The property of the matrix (in general, for any size $n\times n$):
$$\begin{vmatrix}
a+b&c+d \\ e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ e&f
\end{vmatrix}+\begin{vmatrix}
b&d \\ e&f
\end{vmatrix};\\
\begin{vmatrix}
a+b&c \\ d+e&f
\end{vmatrix}
=
\begin{vmatrix}
a&c \\ d&f
\end{vmatrix}+\begin{vmatrix}
b&c \\ e&f
\end{vmatrix};$$
Hence:
$$\begin{vmatrix}
1&3&4 \\ 5&2&(a+2) \\6&-2&3
\end{vmatrix}
=\begin{vmatrix}
1&3&c+(4-c) \\ 5&2&a+2 \ \ \ \ \ \ \ \ \ \  \\6&-2&d+(3-d)
\end{vmatrix}=\\
\begin{vmatrix}
1&3&c \\ 5&2&a \\6&-2&d
\end{vmatrix}+\begin{vmatrix}
1&3&4-c \\ 5&2&2 \\6&-2&3-d
\end{vmatrix}
$$
For easy calculation of determinant you can create as many zeros as possible (especially in rows/columns).
A: +40 comes from the cofactor of the element A(2,3). It is calculated as:
$$C_{2,3} = (-1)^{2+3} \begin{vmatrix}
1 & 3\\
6 & -2
\end{vmatrix} = 20$$
When multiplied with $2$, it becomes 40.
A: The determinant, is after all a multilinear map on the column / row vectors. 
Therefore,
$$
\det\begin{pmatrix}
1&3&4 \\ 5&2&(a+2) \\6&-2&3
\end{pmatrix}
=
\det\begin{pmatrix}
1&3&4 \\ 5&2&a \\6&-2&3
\end{pmatrix}
+\det\begin{pmatrix}
1&3&0 \\ 5&2&2 \\6&-2&0
\end{pmatrix}
$$


*

*The determinant of the second matrix is?

*If I changed $2$ to $3$ or $4$ or $b$, the answer would be?
