Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Appendix C - C - Exercise Set - Page 686: 31

Answer

$$x=-2,y=3,z=-1$$

Work Step by Step

We solve the given system of equations using Cramer's Rule. To do this, we turn the system into two matrices. We then create x, y, and z matrices by replacing the values in the original matrix with corresponding columns in the answer matrix. We then find determinants to solve. Doing this, we find: $$ M=\begin{pmatrix}2&2&1\\ -1&1&2\\ 1&2&4\end{pmatrix} \\ M_2 =\begin{pmatrix}1\\ 3\\ 0\end{pmatrix} $$ So: $$M_x=\begin{pmatrix}1&2&1\\ 3&1&2\\ 0&2&4\end{pmatrix} \\ M_y=\begin{pmatrix}2&1&1\\ -1&3&2\\ 1&0&4\end{pmatrix} \\ M_z=\begin{pmatrix}2&2&1\\ -1&1&3\\ 1&2&0\end{pmatrix} $$ Thus: $$ x=\frac{D_x}{D}=\frac{-18}{9} =-2 \\ y=\frac{D_y}{D}=\frac{27}{9}=3 \\ z=\frac{D_z}{D}=\frac{-9}{9} =-1 $$
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