Question

Find the standard matrix of the linear transformation T. T: R^2->R^2 rotates points (about the origin) through 7/4π radians (with counterclockwise) rotation for a positive angle.

Answer 1

Answer:

Step-by-step explanation:

The standard matrix A for a transformation where $T:R^2\rightarrow R^2$ is a transformation that rotates each point in $R^2$ about the origin through an angle $\theta$

$e_1=\left[\begin{array}{ccc}1\\0\end{array}\right] =\left[\begin{array}{ccc}\cos\theta\\\sin\theta\end{array}\right] \\\\e_2=\left[\begin{array}{ccc}0\\1\end{array}\right] =\left[\begin{array}{ccc}-\sin\theta\\-\cos\theta\end{array}\right]$

Therefore the rotational transformation for $\frac{7}{4\pi}$

$\left[\begin{array}{ccc}\cos\frac{7}{4\pi}&-\sin\frac{7}{4\pi}\\\\\sin\frac{7}{4\pi}&\cos\frac{7}{4\pi}\end{array}\right]$

Answer 2

In the given scenario, A  is the  standard matrix for just a transformation wherein  $T : \mathbb{R}^2 \to \mathbb{R}^2$  is a transformation that rotates every point of $\mathbb{R}^2$ about origin thru an angle $\varphi$:

$e_1=\left[\begin{array}{cc}1\\ 0\\\end{array}\right] = \left[\begin{array}{cc} \cos \varphi \\ \sin \varphi\\\end{array}\right] \\\\\\e_2=\left[\begin{array}{cc}0\\ 1\\\end{array}\right] = \left[\begin{array}{cc} - \sin \varphi \\ \cos \varphi\\\end{array}\right]$

Therefore the transformation of the rotational for $\frac{7 \pi}{4}$ is:

$\to \left[\begin{array}{cc} \cos \frac{7 \pi}{4} & - \sin \frac{7 \pi}{4}\\ \sin \frac{7 \pi}{4} &\cos \frac{7 \pi}{4} \\\end{array}\right] = \left[\begin{array}{cc}\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}\\ - \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2} \\\end{array}\right]$

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