College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 5 - Section 5.2 - Matrix Solution of Linear Systems - 5.2 Exercises - Page 499: 2

Answer

$\begin{bmatrix}1&-4\\0&28\end{bmatrix}$

Work Step by Step

Given matrix is $\begin{bmatrix}1&-4\\7&0\end{bmatrix}$ Row transformation is $R_{2}=-7R_{1}+R_{2}$ The changed matrix is $\begin{bmatrix}1&-4\\-7(1)+7&-7(-4)+0\end{bmatrix}=\begin{bmatrix}1&-4\\0&28\end{bmatrix}$
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