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: 49

Answer

$A= 28$ square units

Work Step by Step

$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{array}\right|$ = see p.645... $=a_{1}b_{2}c_{3}+b_{1}c_{2}a_{3}+c_{1}a_{2}b_{3}-a_{3}b_{2}c_{1}-b_{3}c_{2}a_{1}-c_{3}a_{2}b_{1}$ --------------------- Using the given formula with $(x_{1},y_{1})=(3,-5)$ $(x_{2},y_{2})=(2,6)$ $(x_{3},y_{3})=(-3,5)$ $\left|\begin{array}{lll} 3 & -5 & 1\\ 2 & 6 & 1\\ -3 & 5 & 1 \end{array}\right|=18+15+10-(-18)-15-(-10)$ $=71-15=56$ $A=\displaystyle \pm\frac{1}{2}(56)$ $A= 28$ square units
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