Answer
$\det EA=(\det E)(\det A)$$\star$
Work Step by Step
$(\det E)(\det A)=(k-0)(ad-bc)=kad-kbc$
$\det EA=\begin{vmatrix}a&b\\kc&kd\end{vmatrix}=kad-kbc$
NB: $E=\begin{bmatrix}1&0\\0&k\end{bmatrix}$ is a scaling matrix.
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