Answer
$\det EA=(\det E)(\det A)$$\star$
Work Step by Step
$(\det E)(\det A)=(0-1)(ad-bc)=bc-ad$
$\det EA=\begin{vmatrix}c&d\\a&b\end{vmatrix}=bc-ad$
NB: $E=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ is a row exchange matrix.
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