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. 
 A: Note that the matrix $kA$ has elements $[kA]_{ij}=kA_{ij}$, where $A_{ij}$ are the elements of $A$. If we were to calculate the determinant expression formula, each term has the factor $k$ appearing $n$ times, where $n$ is the dimension of the matrix. You can factor these out from the entire expression, and you're left with something proportional to $\det A$.
Example:
\begin{align}
\det \left(k\begin{bmatrix}
A_{11} & A_{12}\\ 
A_{21} & A_{22}\\  
\end{bmatrix}\right)
=& \det \begin{bmatrix}
kA_{11} & kA_{12}\\ 
kA_{21} & kA_{22}\\  
\end{bmatrix}\\
= & kA_{11}kA_{22}-kA_{21}kA_{12}\\
= & k^2(A_{11}A_{22}-A_{21}A_{12})\\
=& k^2\det A
\end{align}
A: A mnemonic using more advanced stuff, just to recover it from the few determinant rules I manage to remember is:
$Det(kA)=Det(kI_nA)=Det(kI_n)Det(A)=Det\left( \begin{bmatrix}
k & 0 & ... & 0 \\
0 & k & ... & 0 \\
... & ... & ... & ...\\
0 & 0 & ... & k
\end{bmatrix} \right)  Det(A) = k^n Det(A)$
Where I have used $Det(AB)=Det(A)Det(B)$ and a formula for the determinant of diagonal matrices. 
A: In the most simplest of case, say a matrix $A$ which is a $2 \times 2$ matrix, multiplying it by a constant $k$ gives the following general setup.
\begin{equation}
k \begin{pmatrix}
  a & b \\
  c & d 
 \end{pmatrix}
=
\begin{pmatrix}
  ka & kb \\
  kc & kd 
 \end{pmatrix}
\end{equation}
The determinant is therefore (writing $B=kA$);
\begin{eqnarray}
\text{det}B &=& (ka)(kd)-(kb)(kc)\\
            &=& k^{2}ad-k^{2}bc \\
            &=& k^{2}(ad-bc) \\
            &=& k^{2} \text{det}A
\end{eqnarray}
Notice the exponent on $k$ is of order $2$ for the $2 \times 2$ case; a conjecture is that that is would be $3$ for the $3 \times 3$ case. It is now a case of trying to expand that notion to the $n \times n$ case.
A: 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
$$\operatorname{det}(kA)=\sum_{\sigma\in S_n}sign(\sigma)\Pi_{i=1}^nka_{i,\sigma_i}$$
$$=k^n\operatorname{det}(A)$$
