Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 903: 27

Answer

a) The system represented is $\begin{align} & W+2X+5Y+5Z=-3 \\ & W+X+3Y+4Z=-1 \\ & W-X-Y+2Z=3 \end{align}$ b) The complete solution of the system is $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.

Work Step by Step

(a) The system represented by A can be written by multiplication of $\left[ A \right]\times \left[ X \right]=\left[ 0 \right]$ Where A is augmented matrix and X is variable matrix. So, $\left( \begin{matrix} 1 & 2 & 5 & 5 & -3 \\ 1 & 1 & 3 & 4 & -1 \\ 1 & -1 & -1 & 2 & 3 \\ \end{matrix} \right)\times \left( \begin{matrix} W \\ X \\ Y \\ Z \\ 1 \\ \end{matrix} \right)=\left( \begin{align} & 0 \\ & 0 \\ & 0 \\ & 0 \\ & 0 \\ \end{align} \right)$ So, $\begin{align} & W+2X+5Y+5Z=-3 \\ & W+X+3Y+4Z=-1 \\ & W-X-Y+2Z=3 \end{align}$ Hence, the represented system is thus found. (b) Now, $\text{rref}\left( A \right)=\left[ \begin{matrix} 1 & 0 & 1 & 3 & 1 \\ 0 & 1 & 2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \\ \end{matrix} \right]$ So, $\begin{align} & X+2Y+Z=-2 \\ & W+Y+3Z=1 \end{align}$ Now find $ W $ and $ X $ in terms of $ Y $ and $ Z $. So, $\begin{align} & W+Y+3Z=1 \\ & W=1-Y-3Z \end{align}$ And, $\begin{align} & X+2Y+Z=-2 \\ & X=-2-2Y-Z \end{align}$ Hence, the complete solution of the system is $\left\{ \left( 1-Y-3Z,-2-2Y-Z,Y,Z \right) \right\}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.