## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 1 - Systems of Linear Equations - Review Exercises: 32

#### Answer

$x= \frac{3}{2}-2t$ $y=1+2t$ $z=t$

#### Work Step by Step

Write the augmented matrix of the system of linear equations. $\begin{bmatrix} 2 & 1 & 2 & 4\\ 2 & 2 &0 & 5\\ 2 & -1 & 6 & 2 \end{bmatrix}$ Divide the first row by 2. $\begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 2 & 2 &0 & 5\\ 2 & -1 & 6 & 2 \end{bmatrix}$ Add -2 times the 1st row to the 2nd row to produce a new 2nd row. Add -2 times the 1st row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 0 & 1 &-2 & 1\\ 0 & -2 & 4 & -2 \end{bmatrix}$ Add 2 times the 2nd row to the 3rd row to produce a new 3rd row. $\begin{bmatrix} 1 & \frac{1}{2} & 1 & 2\\ 0 & 1 &-2 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}$ Use the parameter t to represent z(where t is any real number) and substitute further for x and y. $z=t$ $y=1+2t$ $x= \frac{3}{2}-2t$

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