## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 8 - Mid-Chapter Check Point - Page 920: 2

#### Answer

The system of equations \begin{align} & 2x+4y+5z=2 \\ & x+y+2z=1 \\ & 3x+5y+7z=4 \end{align} has no solution.

#### Work Step by Step

Consider the following system of equations: \begin{align} & 2x+4y+5z=2 \\ & x+y+2z=1 \\ & 3x+5y+7z=4 \end{align} Express the above system in the form of a matrix as follows: $\left[ \begin{matrix} 2 & 4 & 5 & 2 \\ 1 & 1 & 2 & 1 \\ 3 & 5 & 7 & 4 \\ \end{matrix} \right]$ Using the elementary row operations we will obtain the echelon form of a matrix: ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & 2 & \frac{5}{2} & 1 \\ 1 & 1 & 2 & 1 \\ 3 & 5 & 7 & 4 \\ \end{matrix} \right]$ ${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},\text{ }{{\text{R}}_{3}}\to {{R}_{3}}+\left( -3 \right){{R}_{1}}$ gives, $\left[ \begin{matrix} 1 & 2 & \frac{5}{2} & 1 \\ 0 & -1 & -\frac{1}{2} & 0 \\ 0 & -1 & -\frac{1}{2} & 1 \\ \end{matrix} \right]$ ${{R}_{2}}\to \frac{1}{-1}{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 2 & \frac{5}{2} & 1 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & -1 & -\frac{1}{2} & 1 \\ \end{matrix} \right]$ ${{R}_{3}}\to {{R}_{3}}+1{{R}_{2}}$ gives, $\left[ \begin{matrix} 1 & 2 & \frac{5}{2} & 1 \\ 0 & -1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix} \right]$ Since, $0{\lt}1$, this implies that the system is inconsistent, and has no solution.

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