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) ]