## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 10 - Section 10.3 - Matrices and Systems of Linear Equations - 10.3 Exercises - Page 710: 42

#### Answer

( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $\ \ t\in \mathbb{R}$

#### Work Step by Step

Write the augmented matrix and, using row transformations, arrive at the row-reduced echelon form. $\left[\begin{array}{llll} 1 & -2 & 5 & 3\\ -2 & 6 & -11 & 1\\ 3 & -16 & 20 & -26 \end{array}\right] \ \ \begin{array}{l} .\\ +2R_{2}.\\ -3R_{1}. \end{array}$ $\left[\begin{array}{llll} 1 & -2 & 5 & 3\\ 0 & 2 & -1 & 7\\ 0 & -10 & 5 & -35 \end{array}\right] \ \ \begin{array}{l} +R_{2}.\\ \div 2.\\ -5R_{2}. \end{array}$ $\left[\begin{array}{llll} 1 & 0 & 4 & 10\\ 0 & 1 & -1/2 & 7/2\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \begin{array}{l} .\\ .\\ . \end{array}$ Last row has all zeros, so the system is dependent (infinite number of solutions). Taking $w=t\in \mathbb{R}$ (arbitrary), back-substituting: $x=10-4t$ $y=\displaystyle \frac{7}{2}+\frac{1}{2}t$ Solution set: $\{$( $10-4t$ , $\displaystyle \frac{7}{2}+\frac{1}{2}t$ , $t$ ), $t\in \mathbb{R} \}$

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