College Algebra (10th Edition)

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

Chapter 8 - Section 8.4 - Matrix Algebra - 8.4 Assess Your Understanding - Page 598: 21

Answer

$\left[\begin{array}{ll} 25 & -9\\ 4 & 20 \end{array}\right]$

Work Step by Step

We found in exercise 15: $ AC =\left[\begin{array}{ll} 28 & -9\\ 4 & 23 \end{array}\right],\qquad$ a 2 by 2 matrix. $3I_{2}=3\left[\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 3 & 0\\ 0 & 3 \end{array}\right]\qquad $also a 2 by 2 matrix, so the difference is defined. $AC-3I=\left[\begin{array}{ll} 28-3 & -9-0\\ 4-0 & 23-3 \end{array}\right]=\left[\begin{array}{ll} 25 & -9\\ 4 & 20 \end{array}\right]$
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