Answer
This scaling of row 2 by a constant $k$ multiplies the determinant by $k$.
Work Step by Step
$\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc=\frac{1}{k}(kad-kbc)=\frac{1}{k}\begin{vmatrix}a&b\\kc&kd\end{vmatrix}$, $k\neq 0$
$k=0\implies\begin{vmatrix}a&b\\kc&kd\end{vmatrix}=k(ad-bc)=0=0\begin{vmatrix}a&b\\c&d\end{vmatrix}$
