Answer
$12$
Work Step by Step
$
\begin{align}
\det A &=-3\begin{vmatrix}1&-2&2\\2&-4&5\\2&0&5\end{vmatrix}
=-3\cdot(2\begin{vmatrix}-2&2\\-4&5\end{vmatrix}+5\begin{vmatrix}1&-2\\2&-4\end{vmatrix})\\
&=-6(-10+8)+15(-4+4)=12
\end{align}
$
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: 10
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