Question

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

Answer 1

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

Answer 2

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

The orthogonal basis L in $\rm R^2$ is given by:

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

$\rm T(u_1)=u_1$

$\rm T(u_2)=0$

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

$\rm Q = \dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}$

Now, matrix A is given by the formula:

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

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

$\rm A = \dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\dfrac{1}{\sqrt{39} }\begin{bmatrix}6 & 3\\ 3 & -6\end{bmatrix}$

Simplify the above expression.

$\rm A = \dfrac{1}{{39} }\begin{bmatrix}6 & 18\\ 18 & 9\end{bmatrix}$

$\rm A = \dfrac{3}{{39} }\begin{bmatrix}2 & 6\\ 6 & 3\end{bmatrix}$

$\rm A = \dfrac{1}{{13} }\begin{bmatrix}2 & 6\\ 6 & 3\end{bmatrix}$

Multiply by 1/13 in the matrix.

$\rm A = \begin{bmatrix}\dfrac{2}{13} & \dfrac{6}{13}\\ \dfrac{6}{13} & \dfrac{3}{13}\end{bmatrix}$

For more information, refer to the link given below:

https://brainly.com/question/16850761