MHB Theman123's question at Yahoo Answers (Rotation and reflection)

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The discussion centers on two linear transformations in ℝ²: S, which rotates points clockwise through 60° and then reflects them through the origin, and T, which reflects points through the origin before rotating them. The standard matrix for the clockwise rotation is provided, along with the reflection matrix through the origin. It is noted that the matrix for transformation S is calculated as [S]_{B_c} = BA = -A, while for transformation T, it is [T]_{B_c} = AB = -A. Both transformations result in the same matrix, indicating they are equivalent despite the order of operations.
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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.
 
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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 $[T]_{B_c}=AB=-A$.
 
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