Find the matrix A of the reflection in the line L in R2 that consists of all scalar multiples of the vector.

Answer:

hello your question is incomplete attached below is the complete question

answer : 2/13    6/13

               6/13    3/13

Step-by-step explanation:

Firstly we will calculate the orthogonal basis L in R^2

attached below is the detailed solution

also assuming β to be the standard ordered basis

hence the matrix A of the reflection in the Line L in R2 =

2/13    6/13

6/13    3/13

The matrix A of the reflection in line L is given by  [tex]\rm A = \begin{bmatrix}\dfrac{2}{13} & \dfrac{6}{13}\\ \dfrac{6}{13} & \dfrac{3}{13}\end{bmatrix}[/tex]  and this can be determined by first evaluating the value of [tex]\rm R^2[/tex] by using the given data.

The orthogonal basis L in [tex]\rm R^2[/tex] is given by:

[tex]r = [u_1,u_2] =\left[\dfrac{1}{\sqrt{39} }\binom{6}{3}, \dfrac{1}{\sqrt{39} }\binom{3}{-6}\right][/tex]

[tex]\rm T(u_1)=u_1[/tex]

[tex]\rm T(u_2)=0[/tex]

[tex]\rm [T]_r = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}[/tex]

[tex]\rm Q = \dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}[/tex]

Now, matrix A is given by the formula:

[tex]\rm A=Q[T]_rQ^{-1}[/tex]

Now, substitute the value of known terms in the above equation.

[tex]\rm A = \dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}[/tex]

Simplify the above expression.

[tex]\rm A = \dfrac{1}{{39} }\begin{bmatrix}6 & 18\\ 18 & 9\end{bmatrix}[/tex]

[tex]\rm A = \dfrac{3}{{39} }\begin{bmatrix}2 & 6\\ 6 & 3\end{bmatrix}[/tex]

[tex]\rm A = \dfrac{1}{{13} }\begin{bmatrix}2 & 6\\ 6 & 3\end{bmatrix}[/tex]

Multiply by 1/13 in the matrix.

[tex]\rm A = \begin{bmatrix}\dfrac{2}{13} & \dfrac{6}{13}\\ \dfrac{6}{13} & \dfrac{3}{13}\end{bmatrix}[/tex]

For more information, refer to the link given below:

https://brainly.com/question/16850761

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HELP! WILL GIVE BRANLIEST!
(Question 12)

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