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theman123's question at Yahoo! Answers (Rotation and reflection)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Let S:ℝ2→ℝ2 be the linear transformation that first rotates points clockwise through 60∘ and then reflects points through the origin.
The standard matrix of S is

Let T:ℝ2→ℝ2 be the linear transformation that first reflects points through the origin and then rotates points clockwise through 60∘.
Here is a link to the question:

Let S:

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello theman123,

According to a well-known property, the matrix of the linear transformation that rotates points clockwise through $60^0$ with respect to the canonical basis $B_c$ of $\mathbb{R}^2$ is $$A=\begin{bmatrix}{\cos (-60^{0})}&{-\sin (-60^{0})}\\{\sin (-60^{0})}&{\;\;\cos (-60^{0})}\end{bmatrix}=\begin{bmatrix}{\;\; 1/2}&{\sqrt{3}/2}\\{-\sqrt{3}/2}&{1/2}\end{bmatrix}$$ and the matrix of the reflection through the origin with respect to $B_c$ is: $$B=\begin{bmatrix}{\cos (180^{0})}&{-\sin (180^{0})}\\{\sin (180^{0})}&{\;\;\cos (180^{0})}\end{bmatrix}=\begin{bmatrix}{-1}&{\;\;0}\\{\;\;0}&{-1}\end{bmatrix}=-I$$ So, $_{B_c}=BA=-A$ and $[T]_{B_c}=AB=-A$.