Answer
$k=5$
Work Step by Step
$AB=\begin{bmatrix}23&-10+5k\\-9&15+k\end{bmatrix}$
$BA=\begin{bmatrix}23&15\\6-3k&15+k\end{bmatrix}$
$AB=BA\implies -10+5k=15$
AND $-9=6-3k$
AND $15+k=15+k$,
all of which are solved by $k=5$.
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 2 - Matrix Algebra - 2.1 Exercises - Page 102: 9
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