Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.4 Matrix Algebra - 10.4 Assess Your Understanding - Page 777: 48


$x=\dfrac{-1}{2}, y=-3 \text{ or } \left(\dfrac{-1}{2},3\right)$

Work Step by Step

We will write the system $\left\{\begin{array}{r}{-4x+y=5}\\{6x-2y=-9}\end{array}\right.$ in matrix form as: $AX=B$ where, $X=\left[\begin{array}{l}x\\y \end{array}\right]$ We have: $A=\left[\begin{array}{ll}{-4}&{1}\\{6}&{-2}\end{array}\right]$, and its inverse is: $A^{-1}=\left[\begin{array}{rr}{-1}&{-1/2}\\{-3}&{-2}\end{array}\right]$ Thus, the solution of the given matrix can be expressed as: $X=A^{-1}B=\left[\begin{array}{rr}{-1}&{-1/2}\\{-3}&{-2}\end{array}\right]\left[\begin{array}{l} 5\\-9\end{array}\right]$ $\left[\begin{array}{l} x\\y \end{array}\right]=\left[\begin{array}{l} -5+\dfrac{9}{2}\\ -15+18 \end{array}\right]=\left[\begin{array}{l} \dfrac{-1}{2}\\3\end{array}\right]$ So, our solution is:: $(x, y)=\left(\dfrac{-1}{2},3\right)$
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