## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 903: 16

#### Answer

The complete solution is $\left\{ 2,\ \frac{z}{2}-1,\ z \right\}$.

#### Work Step by Step

We follow the following steps. \begin{align} & 3x+2y-z=5 \\ & x+2y-z=1 \end{align} Now converting it into an augmented matrix, we get, \left[ \left. \begin{align} & \begin{matrix} 3 & 2 & -1 \\ \end{matrix} \\ & \begin{matrix} 1 & 2 & -1 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 5 \\ 1 \\ \end{matrix} \right] \left[ \left. \begin{align} & \begin{matrix} 3 & 2 & -1 \\ \end{matrix} \\ & \begin{matrix} 0 & 4 & -2 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 5 \\ -2 \\ \end{matrix} \right] $3{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ Now we will express this matrix into equations as below:. \begin{align} & 3x+2y-z=5 \\ & 4y-2z=-2 \end{align} Further we will solve the equations for values of $x,y$ and $z$. \begin{align} & x=2 \\ & y=\frac{z}{2}-1 \\ & z=z \end{align} Hence, the solution is: $\left\{ 2,\ \frac{z}{2}-1,\ z \right\}$

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