Answer
$\det EA=(\det E)(\det A)$$\star$
Work Step by Step
$(\det E)(\det A)=(1-0)(ad-bc)=ad-bc$
$\det EA=\begin{vmatrix}a+kc&b+kd\\c&d\end{vmatrix}=(a+kc)d-(b+kd)c=ad+kcd-bc-kcd=ad-bc$
Linear Algebra and Its Applications (5th Edition)
by Lay, David C.; Lay, Steven R.; McDonald, Judi J.
Published by
Pearson
ISBN 10:
032198238X
ISBN 13:
978-0-32198-238-4
Chapter 3 - Determinants - 3.1 Exercises - Page 170: 33
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