Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 3 - Section 3.4 - Matrix Solutions to Linear Systems - Exercise Set - Page 228: 1

Answer

The answer is $\left(\begin{array}{cc|c} 1 & -\frac{3}{2} & 5 \\ 2 & 2 & 5 \\ \end{array}\right)$

Work Step by Step

The given matrix form is $\left(\begin{array}{cc|c} 2 & 2 & 5 \\ 1 & -\frac{3}{2} & 5 \\ \end{array}\right)$ Perform $R_1\leftrightarrow R_2$. Interchange row 1 with row 2 as shown below. The new matrix will be $\left(\begin{array}{cc|c} 1 & -\frac{3}{2} & 5 \\ 2 & 2 & 5 \\ \end{array}\right)$
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