Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.9 Exercises - Page 79: 8

Answer

$T=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$

Work Step by Step

We are looking for a $2 \times 2$ matrix that reflects points through the horizontal axis and then reflects points through the line $x_{2}=x_{1}$ First the points need to be reflected through the horizontal $x_{1}$ -axis. This then means that the $x_{1}$ -coordinate remains unaffected and the $x_{2}$ -coordinate changes sign: \[ R_{1}=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right] \] Note: if you multiply this matrix by the vector $\left(x_{1}, x_{2}\right)^{T}$ to the right, then you get the vector $\left(x_{1},-x_{2}\right)^{T}$ Next the points need to be reflected through the line $x_{2}=x_{1}$. The $x_{1}$ -coordinate and $x_{2}$ -coordinate then interchange. \[ R_{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \] Obseeve: if you multiply this matrix by the vector $\left(x_{1}, x_{2}\right)^{T}$ to the right, then you get the vector $\left(x_{2}, x_{1}\right)^{T}$ The combination of the reflections is then the product of the reflection matrices, with the first reflection matrix to the right. \[ T=R_{2} \times R_{1}=\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \times\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \] Observe: if you multiply this matrix by the vector $\left(x_{1}, x_{2}\right)^{T}$ to the right, then you obtain the vector $\left(-x_{2}, x_{1}\right)^{T}$
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