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 10.3 Systems of Linear Equations: Determinants - 10.3 Assess Your Understanding - Page 761: 23


We cannot apply Cramer's rule.

Work Step by Step

Cramer's rule states that $a x+b y=p \\ cx+dy=q$ $\triangle=\left|\begin{array}{ll}{a}&{b}\\{c}&{d}\end{array}\right|, \triangle_{1}=\left|\begin{array}{ll}{p}&{b}\\{q}&{d}\end{array}\right|, \triangle_{2}=\left|\begin{array}{ll}{a}&{p}\\{c}&{q}\end{array}\right|$; $ x=\dfrac{\triangle_1}{\triangle}; y=\dfrac{\triangle_2}{\triangle} (D\displaystyle \neq 0)$ From the given system of equations, we have: $ \left[\begin{array}{ll}a & b\\c & d \end{array}\right]=\left[\begin{array}{ll} 3 & -2\\ 6 & 4\end{array}\right],\quad \left[\begin{array}{l} p\\q \end{array}\right]=\left[\begin{array}{l}4\\0\end{array}\right]$ But $\triangle = \left|\begin{array}{ll} 3 & -2\\ 6 & 4 \end{array}\right|=12-12=0$ We see that determinant of the given system is $0$. So, we cannot apply Cramer's rule to find the values of $x$ and $y$.
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