## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Review Exercises - Page 950: 9

#### Answer

The solution is, $x=2t+4,y=t+1,z=t$

#### Work Step by Step

Consider the given system of equations \begin{align} & x-3y+z=1 \\ & -2x+y+3z=-7 \\ & x-4y+2z=0 \end{align} Therefore, in matrix form the system of equations can be written as below: $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 1 & -3 & 1 \\ -2 & 1 & 3 \\ 1 & -4 & 2 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 1 \\ -7 \\ 0 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & -3 & 1 & 1 \\ -2 & 1 & 3 & -7 \\ 1 & -4 & 2 & 0 \\ \end{array} \right]$ Apply elementary row operation on $A$ to convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}+2{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-{{R}_{1}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & -3 & 1 & 1 \\ 0 & -5 & 5 & -5 \\ 0 & -1 & 1 & -1 \\ \end{array} \right]$ Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}-\frac{1}{5}{{R}_{2}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & -3 & 1 & 1 \\ 0 & -5 & 5 & -5 \\ 0 & 0 & 0 & 0 \\ \end{array} \right]$ Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $A$ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=2$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.