College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Section 8.3 - Systems of Linear Equations: Determinants - 8.3 Assess Your Understanding - Page 582: 11

Answer

$10$

Work Step by Step

2 by 2 determinant $D=\left|\begin{array}{ll} a & b\\ c & d \end{array}\right|=ad-bc$ 3 by 3 determinant (cofactors of first row): $\left|\begin{array}{lll} {a_{11}}&{a_{12}}&{a_{13}}\\ {a_{21}}&{a_{22}}&{a_{23}}\\ {a_{31}}&{a_{32}}&{a_{33}}\end{array}\right|=a_{11}\left|\begin{array}{cc} {a_{22}}&{a_{23}}\\ {a_{32}}&{a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{cc} {a_{21}}&{a_{23}}\\ {a_{31}}&{a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{cc} {a_{21}}&{a_{22}}\\{a_{31}}&{a_{32}}\end{array}\right|$ --- $D=3\left|\begin{array}{ll} -1 & 5\\ 2 & -2 \end{array}\right|-4\left|\begin{array}{ll} 1 & 5\\ 1 & -2 \end{array}\right|+2\left|\begin{array}{ll} 1 & -1\\ 1 & 2 \end{array}\right|$ $=3[(-1)(-2)-2(5)]-4[1(-2)-1(5)]$ $=3(-8)-4(-7)+2(3)$ $=-24+28+6$ $=10$
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