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: 39

Answer

$-173$

Work Step by Step

We are given the determinant: $D=\begin{vmatrix}-2&-3&3&5\\1&-4&0&0\\1&2&2&-3\\2&0&1&1\end{vmatrix}$ In order to compute the determinant we expand it along the second row because it has 2 zeros: $D=-1D_{21}+(-4)D_{12}-0D_{13}+0D_{14}=-D_{21}4D_{22}$ $=-\begin{vmatrix}-3&3&5\\2&2&-3\\0&1&1\end{vmatrix}-4\begin{vmatrix}-2&3&5\\1&2&-3\\2&1&1\end{vmatrix}$ $=-\left(-0\begin{vmatrix}3&5\\2&-3\end{vmatrix}-1\begin{vmatrix}-3&5\\2&-3\end{vmatrix}+1\begin{vmatrix}-3&3\\2&2\end{vmatrix}\right)-4\left(-2\begin{vmatrix}2&-3\\1&1\end{vmatrix}-3\begin{vmatrix}1&-3\\2&1\end{vmatrix}+5\begin{vmatrix}1&2\\2&1\end{vmatrix}\right)$ $=-[0-1(9-10)+1(-6-6)]-4[-2(2+3)-3(1+6)+5(1-4)]$ $=11-184=-173$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.