Answer
False
Work Step by Step
Given $nullspace (A) =\{0\}$. Thus, $A$ is invertible.
Since $E$ is an $n\times n$ elementary matrx, $E$ is invertible.
$EA$ is invertible. Hence $\det (EA)\ne 0$ The statement is false.
Differential Equations and Linear Algebra (4th Edition)
by Goode, Stephen W.; Annin, Scott A.
Published by
Pearson
ISBN 10:
0-32196-467-5
ISBN 13:
978-0-32196-467-0
Chapter 4 - Vector Spaces - 4.10 Invertible Matrix Theorem II - True-False Review - Page 332: f
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