Determinant of a Matrix Proof: $\;\det(qA) = q^n(\det A)$ I am required to show that:
$\det(qA) = q^n(\det A)$, where $A$ is a real $n\times n$ Matrix, and $q$ is a constant 
I believe that this claim is true after doing few examples. However, but I do not know how to start the proof. 

HINT:
Use the definition
$$
\det(a_{ij})=\sum_{\pi}s(\pi)a_{1,\pi(1)}\cdots a_{n,\pi(n)}
$$
where the sum is extended to all permutations of the set $\{1,...,n\}$ and $s(\pi)=\pm1$ is the sign of the permutation $\pi$.
What happens if every $a_{ij}$ is replaced by $qa_{ij}$?
