Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.6 Multiplying Matrices - 3.6 Exercises - Skill Practice - Page 200: 27

Answer

$(D+E)D=\begin{bmatrix} -2&4&0\\ 5&15 & 8\\ -16 & 17 &36 \end{bmatrix}$

Work Step by Step

$(D+E)D=(\begin{bmatrix} 1&3&2\\ -3& 1 & 4\\ 2 & 1 & -2 \end{bmatrix}+\begin{bmatrix} -3&1&4\\ 7& 0 & -2\\ 3 & 4 & -1 \end{bmatrix})\begin{bmatrix} 1&3&2\\ -3& 1 & 4\\ 2 & 1 & -2 \end{bmatrix}$ $=\begin{bmatrix} -2&4&6\\ 4& 1 & 2\\ 5 &5 & -3 \end{bmatrix}.\begin{bmatrix} 1&3&2\\ -3& 1 & 4\\ 2& 1 & -2 \end{bmatrix}$ $=\begin{bmatrix} -2-12+12 & -6+4+6 & -4 + 16 -12\\ 4 -3+4& 12+1+2& 8+4-4\\ 5-15-6 & 15+5-3 & 10+20+6 \end{bmatrix}$ $=\begin{bmatrix} -2&4&0\\ 5&15 & 8\\ -16 & 17 &36 \end{bmatrix}$
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