College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.1 - Page 599: 18

Answer

$\left[\begin{array}{llllll} 1 & -5 & 2 & -2 & | & 4\\ 0 & 1 & -3 & -1 & | & 0\\ 0 & 15 & -4 & 5 & | & -6\\ 0 & -19 & 12 & -6 & | & 13 \end{array}\right]$

Work Step by Step

The elements of the third row will be replaced with $-3(1)+3=0,$ $-3(-5)+0=15,$ $-3(2)+2=-4,$ $-3(-2)+(-1)=5,$ $-3(4)+6=-6$, to become $0\ \ 15\ \ -4\ \ 5\ \ |\ \ -6.$ The elements of the fourth row will be replaced with $4(1)+(-4)=0,$ $4(-5)+1=-19,$ $4(2)+4=12,$ $4(-2)+2=-6,$ $4(4)+(-3)=13$ , to become $0\ \ -19\ \ 12\ \ -6\ \ |\ \ 13$ resulting in $\left[\begin{array}{llllll} 1 & -5 & 2 & -2 & | & 4\\ 0 & 1 & -3 & -1 & | & 0\\ 0 & 15 & -4 & 5 & | & -6\\ 0 & -19 & 12 & -6 & | & 13 \end{array}\right]$
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