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The matrix of Linear Transformation $\text{T}\left(\stackrel{鈫�}{x}\right)=B\left(A\stackrel{鈫�}{x}\right),鈥�鈥�鈭赌\stackrel{鈫�}{x}鈭�{\mathrm{鈩�}}^2$ is known as the product of matrices $B\text{and}A鈬�BA$, can be given by$\text{T}\left(\stackrel{鈫�}{x}\right)=B\left(A\stackrel{鈫�}{x}\right)=\left(BA\right)\stackrel{鈫�}{x},鈥�鈥�鈭赌\stackrel{鈫�}{x}鈭�{\mathrm{鈩�}}^2$:

Given, the lines P and Q in ${\mathrm{鈩�}}^2$ are shown in figure below:

The given linear transformation is:$T\left(\stackrel{鈫�}{x}\right)={\text{ref}}_Q\left({\text{ref}}_P\left(\stackrel{鈫�}{x}\right)\right)$ .

Initially the reflection of $\stackrel{鈫�}{x}$ is along P which can be shown in below graph

And, the reflection of ${\text{ref}}_P\left(\stackrel{鈫�}{x}\right)$ is along Q which can be shown in below graph:

This sketch is representing $T\left(\stackrel{鈫�}{x}\right)$.

From above sketch, the angle between $\stackrel{鈫�}{x}\text{and}T\left(\stackrel{鈫�}{x}\right)$ will be:

$胃+胃+\left(30掳-胃\right)+\left(30掳-胃\right)=60掳$

Also, the vectors $\stackrel{鈫�}{x}\text{and}T\left(\stackrel{鈫�}{x}\right)$ are found to be equal in length.

Since, the angle between $\stackrel{鈫�}{x}\text{and}T\left(\stackrel{鈫�}{x}\right)$ is 60⁰ .

Thus, it can be said that the transformation of $T\left(\stackrel{鈫�}{x}\right)$ is the result of rotation of $\stackrel{鈫�}{x}$ with an angle of 60⁰ in counter-clockwise direction.

As we know, the rotation matrix of linearly transformed vector is given by:

$A=\left[\begin{array}{cc}\cos 胃& -\sin 胃\\ \sin 胃& \cos 胃\end{array}\right]$

Since, the transformation of $T\left(\stackrel{鈫�}{x}\right)$ is the result of rotation of $\stackrel{鈫�}{x}$ with an angle of 60⁰ .

Therefore, the matrix of the transformation of $T\left(\stackrel{鈫�}{x}\right)$ will be:

$$\begin{gathered} A=\left[\begin{array}{cc}\cos 60掳& -\sin 60掳\\ \sin 60掳& \cos 60掳\end{array}\right] \\ A=\left[\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right] \end{gathered}$$

Hence, this is the required matrix.

From part (a), the angle between $\stackrel{鈫�}{x}$and Q is 45⁰ . Initially the reflection of $\stackrel{鈫�}{x}$ is along Q and the reflection of ${\text{ref}}_Q\left(\stackrel{鈫�}{x}\right)$ is perpendicular to Q .

Also, the angle between P and Q is 30⁰ and the reflection of along vector P makes angle 120⁰ . This results the reflection vector represented by $L\left(\stackrel{鈫�}{x}\right)={\text{ref}}_P\left({\text{ref}}_Q\left(\stackrel{鈫�}{x}\right)\right)$.

In this case, the result of rotation of $\stackrel{鈫�}{x}$ with an angle of 60⁰ is in clockwise direction.

Therefore, the angle between $\stackrel{鈫�}{x}\text{andL}\left(\stackrel{鈫�}{x}\right)$ is -60⁰ .

Now, the transformation of $\text{L}\left(\stackrel{鈫�}{x}\right)$ is the result of rotation of -60⁰ with an angle of .

Therefore, the matrix of the transformation of $\text{L}\left(\stackrel{鈫�}{x}\right)$ will be:

$$\begin{gathered} A=\left[\begin{array}{cc}\cos \left(-60掳\right)& -\sin \left(-60掳\right)\\ \sin \left(-60掳\right)& \cos \left(-60掳\right)\end{array}\right] \\ A=\left[\begin{array}{cc}\frac{1}{2}& \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right] \end{gathered}$$

Hence, this is the required matrix, where $\text{L}\left(\stackrel{鈫�}{x}\right)$ is found by the rotation of $T\left(\stackrel{鈫�}{x}\right)$.