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: 47

Answer

$x=4$

Work Step by Step

$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{array}\right| =$ $=a_{1}b_{2}c_{3}+b_{1}c_{2}a_{3}+c_{1}a_{2}b_{3}-a_{3}b_{2}c_{1}-b_{3}c_{2}a_{1}-c_{3}a_{2}b_{1}$ --------------------- $\left|\begin{array}{lll} 1 & x & -2\\ 3 & 1 & 1\\ 0 & -2 & 2 \end{array}\right|$ $=1(1)(2)+x(1)(0)+(-2)(3)(-2)-0(1)(-2) - (-2)(1)(1)-2(3)(x)$ $=2+12+2-6x=16-6x$ $16-6x=-8\qquad/-16$ $-6x=-24\qquad/\div(-6)$ $x=4$
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