# Chapter 6 - Summary, Review, and Test - Review Exercises - Page 655: 2

$\left[\begin{array}{lllll} 1 & -1 & 1/2 & | & -1/2\\ 1 & 2 & -1 & | & 2\\ 6 & 4 & 3 & | & 5 \end{array}\right]$

#### Work Step by Step

The elements of the third row will be replaced with $2\displaystyle \rightarrow\quad\frac{1}{2}\cdot 2=1$ $-2\displaystyle \rightarrow\quad\frac{1}{2}\cdot(-2)=-1$ $1\displaystyle \rightarrow\quad\frac{1}{2}\cdot(1)=\frac{1}{2}$ $-1\displaystyle \rightarrow\quad\frac{1}{2}\cdot(-1)=-\frac{1}{2}$ so the new matrix is $\left[\begin{array}{lllll} 1 & -1 & 1/2 & | & -1/2\\ 1 & 2 & -1 & | & 2\\ 6 & 4 & 3 & | & 5 \end{array}\right]$

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