College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Test - Page 636: 8

Answer

See below.

Work Step by Step

$\begin{bmatrix} 1& -1 \\ 0& -4 \\ 3&2 \end{bmatrix}-3\begin{bmatrix} 4& 6 \\ 1& -3\\ -1&8 \end{bmatrix}=\begin{bmatrix} 1& -1 \\ 0& -4 \\ 3&2 \end{bmatrix}-\begin{bmatrix} 12& 18 \\ 3& -9\\ -3&24 \end{bmatrix}=\begin{bmatrix} 1-12& -1-18 \\ 0-3& -4-(-9) \\ 3-(-3)&2-24 \end{bmatrix}=\begin{bmatrix} -11& -19 \\ -3& 5 \\ 6&-22 \end{bmatrix}$
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