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: 34

Answer

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

Work Step by Step

Given $B=\frac{1}{2}(A+A^T)$ $=\frac{1}{2}(\begin{bmatrix} 4 & -1 &0\\ 9 & -2 & 3\\ 2 & 5 & 5 \end{bmatrix}+\begin{bmatrix} 4 & 9 &2\\ -1 & -2 & 5\\ 0 & 3 & 5 \end{bmatrix})$ $=\frac{1}{2}\begin{bmatrix} 8 & 8 &2\\ 8 & -4 & 8\\ 2 & 8 & 10 \end{bmatrix}$ $=\begin{bmatrix} 4 & 4 &1\\ 4 & -2 & 4\\ 1& 4 & 5 \end{bmatrix}$ $C=\frac{1}{2}(A-A^T)$ $=\frac{1}{2}(\begin{bmatrix} 4 & -1 &0\\ 9 & -2 & 3\\ 2 & 5 & 5 \end{bmatrix}-\begin{bmatrix} 4 & 9 &2\\ -1 & -2 & 5\\ 0 & 3 & 5 \end{bmatrix})$ $=\frac{1}{2}\begin{bmatrix} 0 & -10 &-2\\ 10 & 0 & -2\\ 2 & 2 & 0 \end{bmatrix}$ $=\begin{bmatrix} 0 & -5 &-1\\ 5 & 0 & -1\\ 1& 1 & 0 \end{bmatrix}$
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