91Ó°ÊÓ

The matrix of Linear Transformation $\text{T}\left(\stackrel{→}{x}\right)=B\left(A\stackrel{→}{x}\right),\text{ â¶Ä‰}âˆ\P Ä\stackrel{→}{x}{∈}^2$is known as the product of matrices $B\text{and}A⇒BA$, can be given by:

role="math" localid="1664184995265" $\text{T}\left(\stackrel{→}{x}\right)=B\left(A\stackrel{→}{x}\right)=\left(BA\right)\stackrel{→}{x},\text{ â¶Ä‰}âˆ\P Ä\stackrel{→}{x}{∈}^2$

Let, there be a line $L{∈}^3$. Then the orthogonal projection of vector $\stackrel{→}{x}$on the line will be given by:

$\begin{array}{c}{\text{proj}}_L\left(\stackrel{→}{x}\right)=\stackrel{→}{x}\\ =(\stackrel{→}{x}â‹\dots \stackrel{→}{u})\stackrel{→}{u}\end{array}$

Where, $\stackrel{→}{u}$is the unit vector parallel to lineL.

Let, $\stackrel{→}{u}=\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)$

Then, we have:

$\begin{array}{c}{\text{proj}}_L\left(\stackrel{→}{x}\right)=(\stackrel{→}{x}â‹\dots \stackrel{→}{u})\stackrel{→}{u}\\ =(x{u}_1+y{u}_2+z{u}_3)\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)\end{array}$

Now, using the given matrix $B=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$, we have the transformation as follows:

$\begin{array}{c}B\stackrel{→}{x}=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\stackrel{→}{x}\\ =\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{c}x+y+z\\ x+y+z\\ x+y+z\end{array}\right]\\ =[\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z]\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right]\end{array}$

Here, the unit vector obtained is $\stackrel{→}{u}=\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right]$and the line L is $x=y=z$.

Hence, the matrix B represents the orthogonal projection.

Now, the reflection of vector $\stackrel{→}{x}$on the line will be given by:

$\begin{array}{c}{\text{ref}}_L\left(\stackrel{→}{x}\right)=2{\text{proj}}_L\left(\stackrel{→}{x}\right)−\stackrel{→}{x}\\ =2(\stackrel{→}{x}â‹\dots \stackrel{→}{u})\stackrel{→}{u}−\stackrel{→}{x}\end{array}$

Evaluate the above expression as:

$\begin{array}{c}{\text{ref}}_L\left(\stackrel{→}{x}\right)=2(\stackrel{→}{x}â‹\dots \stackrel{→}{u})\stackrel{→}{u}−\stackrel{→}{x}\\ =2(x{u}_1+y{u}_2+z{u}_3)\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\end{array}\right)−\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\end{array}$

Now, using the given matrix $E=\frac{1}{3}\left[\begin{array}{ccc}−1& 2& 2\\ 2& −1& 2\\ 2& 2& −1\end{array}\right]$, we have the transformation as follows:

$\begin{array}{c}B\stackrel{→}{x}=\frac{1}{3}\left[\begin{array}{ccc}−1& 2& 2\\ 2& −1& 2\\ 2& 2& −1\end{array}\right]\stackrel{→}{x}\\ =\frac{1}{3}\left[\begin{array}{c}−x+2y+2z\\ 2x−y+2z\\ 2x+2y−z\end{array}\right]+\left[\begin{array}{c}x\\ y\\ z\end{array}\right]−\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ =\frac{1}{3}\left[\begin{array}{c}2x+2y+2z\\ 2x+2y+2z\\ 2x+2y+2z\end{array}\right]−\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\\ =2[\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z]\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right]−\left[\begin{array}{c}x\\ y\\ z\end{array}\right]\end{array}$

Hence, the matrix E represents the reflection onto the lineL is $x=y=z$.