<s>[INST] Am I misinterpreting this matrix determinant property? I was reading matrix determinant properties from wikipedia.
The property reads
$\det(cA) = c^n \det(A)$ for $n \times n$ matrix.
However I am not able to realize it. What I find is $\det(cA) = c\det(A)$ 
For example, multiplying matrix by 2 and then taking the determinant of the resultant matrix:
$
2\begin{bmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{bmatrix}=
\begin{bmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{bmatrix}
$
and
$
\begin{vmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{vmatrix}=60
$
Now first taking the determinant and then multiplying by 2 yields the same result:
$$
2\begin{vmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{vmatrix}
= 2 \cdot 30 = 60
$$ 
Where I am mistaking?
 [/INST] You have $\det (cB) = \det (cI) \det B $ and you can see from the formula for $\det$ that $\det (cI) = c^n$.
Another way is to notice that $\det$ is a multilinear function of the columns (or rows), that is, we can write
$\det(A) = f(a_1,...,a_n)$ where $a_k$ is the $k$th row of $A$, and $f$ is
linear in each argument separately.
Then $\det(cA) = f(c a_1,...,c a_n) = c f(a_1, c a_2,...,c a_n) = c^2 f(a_1,  a_2,...,c a_n)= \cdots = c^n f(a_1,...,a_n) = c^n \det A$.
Addendum:
You multiplied the matrix incorrectly.
$A=\begin{bmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{bmatrix}$, $2A = \begin{bmatrix}
8 & 10 & 12 \\
12 & 10 & 8  \\
8 & 12 & 10
\end{bmatrix}$.
$\det A = 30, \det (2A) = 240$.
</s>