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 652: 17

Answer

Solution set = $\displaystyle \{(2,\frac{1}{2})\}$

Work Step by Step

$D= $determinant of the coefficient matrix $=\left|\begin{array}{ll} 1 & 2\\ 3 & -4 \end{array}\right|=1(-4)-(2)(3)=-10$ $D_{x}=$ in D, replace the x column with the constants column $=\left|\begin{array}{ll} 3 & 2\\ 4 & -4 \end{array}\right|=-12-8=-20$ $D_{y}=$ in D, replace the x column with the constants column $=\left|\begin{array}{ll} 1 & 3\\ 3 & 4 \end{array}\right|=4-9=-5$ $x=\displaystyle \frac{D_{x}}{D}=\frac{-20}{-10}=2$ $y=\displaystyle \frac{D_{y}}{D}=\frac{-5}{-10}=\frac{1}{2}$ Solution set = $\displaystyle \{(2,\frac{1}{2})\}$
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