Determinant of matrix times a constant. Prove that $\det(kA) = k^n \det(A)$ for and ($n \times n$) matrix. 
I have tried looking at this a couple of ways but can't figure out where to start. It's confusing to me since the equation for a determinant is such a weird summation. 

To elaborate on the above answer, the formula for the determinant is
$$\operatorname{det}(A)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^na_{i,\sigma_i}$$
so
$$\operatorname{det}(kA)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^nka_{i,\sigma_i}$$
$$=k^n\operatorname{det}(A)$$
