This a question that has to do with a Rotation about the origin. It is to be noted that T: [tex]\mathbb{R}^{2}[/tex] → [tex]\mathbb{R}^{2}[/tex] As given in the question. See the explanation below.
First Case:
T (R1, R2) = (+r1, -r2)
Let Bl = {(1,0), (0,1)} Standard Bases
⇒ (T(1, 0) = (+1, 0)
T (0, 1) = (0, -1)
Then matrix is:
A1 = [tex]\begin{bmatrix} +1& 0\\0 & -1\end{bmatrix}[/tex] and A1 R1 = [tex]\begin{bmatrix} r1\\y2\end{bmatrix}[/tex]
Second Case
T (1, 0) = (0, 1)
T (0, 1) = (1, 0)
The Matrix is:
A2 = [tex]\begin{bmatrix} 0& 1\\1 & 0\end{bmatrix}[/tex]
Hence,
TA2 (TA1(r1)) = [tex]\begin{bmatrix} 0& 1\\1 & 0\end{bmatrix}[/tex] [tex]\begin{bmatrix} r1\\r2\end{bmatrix}[/tex] = [tex]\begin{bmatrix} -r2\\r1\end{bmatrix}[/tex]
Thus
(TA2 * TA1) (r) = [tex]\begin{bmatrix} 0& -1\\1 & 0\end{bmatrix}[/tex] [tex]\begin{bmatrix} r1\\r2\end{bmatrix}[/tex]
The standard Matrix is:
[tex]\mathbb{A} = \begin{bmatrix} 0& -1\\1 & 0\end{bmatrix}[/tex]
In this case,
T (0, 0) = (0, 0)
⇒ Transformation is rotation about the origin.
Hence Rotation Matrix is given as:
Tθ = [tex]\begin{bmatrix} Cos \theta & - Sin \theta\\Sin \theta & Cos\theta\end{bmatrix}[/tex] ⇒ θ = π/2
Learn more about Rotation about the Origin:
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