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 136: 19

Answer

See below

Work Step by Step

$\begin{bmatrix} A_1,A_2 \end{bmatrix}=A_1A_2-A_2A_1=\begin{bmatrix} 1 & 0\\0 &1 \end{bmatrix}\begin{bmatrix} 0 &1\\0 &0 \end{bmatrix}-\begin{bmatrix} 0 &1\\0 &0 \end{bmatrix}\begin{bmatrix} 1 &0\\0 &1 \end{bmatrix}=\begin{bmatrix} 0 &1\\0 & 0 \end{bmatrix}-\begin{bmatrix} 0 &1\\0 &0 \end{bmatrix}$ $A_1$ and $A_3$ commute. $\begin{bmatrix} A_1,A_3 \end{bmatrix}=A_1A_3-A_3A_1=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}\begin{bmatrix} 0 &0\\1&0 \end{bmatrix}-\begin{bmatrix} 0 &0 \\1&0 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix}=\begin{bmatrix} 0&0\\1&0 \end{bmatrix}-\begin{bmatrix} 0&0\\1&0 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ $A_1$ and $A_3$ commute. $\begin{bmatrix} A_2,A_3 \end{bmatrix}=A_2A_3-A_3A_2=\begin{bmatrix} 0&1\\0&0\end{bmatrix}\begin{bmatrix} 0&0\\1&0 \end{bmatrix}-\begin{bmatrix} 0&0 \\1&0 \end{bmatrix}\begin{bmatrix} 0&1\\0& 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\0& 0 \end{bmatrix}-\begin{bmatrix} 0 &0\\0&1 \end{bmatrix}=\begin{bmatrix} 1 &0\\0&-1 \end{bmatrix}$ Since $\begin{bmatrix} A_3,A_2 \end{bmatrix}=-\begin{bmatrix} A_2,A_3 \end{bmatrix}=\begin{bmatrix} -1&0\\0&1 \end{bmatrix}\ne 0_2$ Hence, $A_2$ and $A_3$ do not commute
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