Determine the standard matrix D of the linear mapping G: R^2--R^2 that first rotates points clockwise through pi/6 radians and then reflects points through the line x2=x1

To determine the standard matrix D of the linear mapping G, which involves rotating points clockwise through π/6 radians and then reflecting them through the line x_2 = x_1, we can break down the process into two steps:

Step 1: Rotating points clockwise through π/6 radians:

To rotate a point (x, y) clockwise through an angle θ, we use the following formulas:

x' = x*cos(θ) - y*sin(θ)
y' = x*sin(θ) + y*cos(θ)

In this case, θ = π/6. So, we have:

x' = x*cos(π/6) - y*sin(π/6)
y' = x*sin(π/6) + y*cos(π/6)

Simplifying the equations:

x' = (sqrt(3)/2)x - (1/2)y
y' = (1/2)x + (sqrt(3)/2)y

Step 2: Reflecting points through the line x_2 = x_1:

To reflect a point (x, y) through the line x_2 = x_1, we swap the x-coordinate with the y-coordinate. So, the reflection can be written as:

x' = y
y' = x

Combining both transformations, we get:

x'' = (1/2)x' + (sqrt(3)/2)y'
y'' = (sqrt(3)/2)x' - (1/2)y'

Substituting x' and y' from the previous step:

x'' = (1/2)((sqrt(3)/2)x - (1/2)y) + (sqrt(3)/2)((1/2)x + (sqrt(3)/2)y)
y'' = (sqrt(3)/2)((sqrt(3)/2)x - (1/2)y) - (1/2)((1/2)x + (sqrt(3)/2)y)

Simplifying further:

x'' = (sqrt(3)/4)x - (3/4)y + (3/4)x + (sqrt(3)/4)y
y'' = (3/4)x + (sqrt(3)/4)y - (sqrt(3)/4)x + (1/4)y

Combining like terms:

x'' = (3/4)x + (sqrt(3)/4)y
y'' = -(sqrt(3)/4)x + (5/4)y

Therefore, the standard matrix D of the linear mapping G is:

D = [ (3/4) (sqrt(3)/4) ]
[ -(sqrt(3)/4) (5/4) ]