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: 3

Answer

The answer is $\left(\begin{array}{cc|c} 1 & -\frac{4}{3} & 2 \\ 3 & 5 & -2 \\ \end{array}\right)$

Work Step by Step

The given matrix is $\left(\begin{array}{cc|c} -6 & 8 & -12 \\ 3 & 5 & -2 \\ \end{array}\right)$ Perform $ -\frac{1}{6} R_1$ Multiply row 1 by $-\frac{1}{6}$. $\left(\begin{array}{cc|c} -6\cdot (-\frac{1}{6}) & 8\cdot (-\frac{1}{6}) & -12\cdot (-\frac{1}{6}) \\ 3 & 5 & -2 \\ \end{array}\right)$ Simplify. $\left(\begin{array}{cc|c} 1 & -\frac{4}{3} & 2 \\ 3 & 5 & -2 \\ \end{array}\right)$ .
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