Answer
$\det kA=k^{2}\det A$
Work Step by Step
$\det k\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{vmatrix}ka&kb\\kc&kd\end{vmatrix}=(ka\cdot kd)-(kb\cdot kc)=k^{2}ad-k^{2}bc=k^{2}(ad-bc)=k^{2}\det A$
Note that this formula applies only to $2\times 2$ matrices, but the analog for the $n\times n$ case is readily derived by similar calculations.