Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 3 - Determinants - 3.2 Exercises - Page 177: 12

Answer

$6$

Work Step by Step

We solve by reducing the $4\times 4$ determinant to a $3\times 3$ determinant. At that point, there are no easy row exchanges left, so we proceed directly to computing the $3\times 3$ determinant, rather than reducing it further. $\begin{vmatrix}-1&2&3&0\\3&4&3&0\\11&4&6&6\\4&2&4&3\end{vmatrix}=\begin{vmatrix}-1&2&3&0\\0&10&12&0\\0&26&39&6\\0&10&16&3\end{vmatrix}=\begin{vmatrix}-1&2&3&0\\0&10&12&0\\0&6&15&6\\0&0&4&3\end{vmatrix}=-1\begin{vmatrix}10&12&0\\6&15&6\\0&4&3\end{vmatrix}=-1\cdot \left [10\begin{vmatrix}15&6\\4&3\end{vmatrix}-6\begin{vmatrix}12&0\\4&3\end{vmatrix} \right ]=-210+216=6$
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