Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 2 - Matrices and Systems of Linear Equations - 2.2 Matrix Algebra - Problems - Page 137: 35

Answer

$B=\begin{bmatrix} 1 & -1 &5\\ -1 & 2 & 1\\ 5& 1 & 6 \end{bmatrix}$ $B=\begin{bmatrix} 0 & -4 &-2\\ 4 & 0 & 3\\ 2& -3 &0 \end{bmatrix}$

Work Step by Step

Given $B=\frac{1}{2}(A+A^T)$ $=\frac{1}{2}(\begin{bmatrix} 1 & -5 &3\\ 3 & 2 & 4\\ 7 & -2 & 6 \end{bmatrix}+\begin{bmatrix} 1 & 3 &7\\ -5 & 2 & -2\\ 3 & 4 & 6 \end{bmatrix})$ $=\frac{1}{2}\begin{bmatrix} 2 & -2 &10\\ -2 & 4 & 2\\ 10 & 2 & 12 \end{bmatrix}$ $=\begin{bmatrix} 1 & -1 &5\\ -1 & 2 & 1\\ 5& 1 & 6 \end{bmatrix}$ $C=\frac{1}{2}(A-A^T)$ $=\frac{1}{2}(\begin{bmatrix} 1 & -5 &3\\ 3 & 2 & 4\\ 7 & -2 & 6 \end{bmatrix}-\begin{bmatrix} 1 & 3 &7\\ -5 & 2 & -2\\ 3 & 4 & 6 \end{bmatrix})$ $=\frac{1}{2}\begin{bmatrix} 0 & -8 &-4\\ 8 & 0 & 6\\ 4 & -6 & 0 \end{bmatrix}$ $=\begin{bmatrix} 0 & -4 &-2\\ 4 & 0 & 3\\ 2& -3 &0 \end{bmatrix}$
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