College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.5 - Page 653: 40

Answer

$48$

Work Step by Step

We are given the determinant: $D=\begin{vmatrix}1&-3&2&0\\-3&-1&0&-2\\2&1&3&1\\2&0&-2&0\end{vmatrix}$ In order to compute the determinant we expand it along the fourth row because it has 2 zeros: $D=-2D_{41}+0D_{42}-(-2)D_{43}+0D_{44}=-2D_{41}+2D_{43}$ $=-2\begin{vmatrix}-3&2&0\\-1&0&-2\\1&3&1\end{vmatrix}+2\begin{vmatrix}1&-3&0\\-3&-1&-2\\2&1&1\end{vmatrix}$ $=-2\left(-3\begin{vmatrix}0&-2\\3&1\end{vmatrix}-2\begin{vmatrix}-1&-2\\1&1\end{vmatrix}+0\right)+2\left(1\begin{vmatrix}-1&-2\\1&1\end{vmatrix}+3\begin{vmatrix}-3&-2\\2&1\end{vmatrix}+0\right)$ $=-2[-3(0+6)-2(-1+2)]+2[(-1+2)+3(-3+4)]$ $=-2(-20)+2(4)=48$
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