Matrix transformation

**terrorsquid** · November 20th 2011, 08:14 PM

Consider the linear transformation, $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ , for which $T(x)$ is found by rotating the point x through an angle of $\frac{2\pi}{3}$ anticlockwise about the origin.

a) Find the 2 x 2 matrix A for which T(x) = Ax

I wasn't quite sure how $T(x)$ could = Ax and as such I wasn't entirely sure how to answer the question. I just applied the same method I had been using for all the previous question to get:

$\left(\begin{matrix} cos(\frac{2\pi}{3}) &-sin(\frac{2\pi}{3}) \\sin(\frac{2\pi}{3}) & cos(\frac{2\pi}{3})\end{matrix}\right)$

Is this the correct answer for matrix A? I thought this would be $T(x)$ ? What is the x value that A is multiplied by?

b) Write down the transformation that returns the point x to its original position.

Since we rotated everything $\frac{2\pi}{3}$ counter-clockwise, I would presume that to reverse that transformation we would rotate everything by $-\frac{2\pi}{3}$ to get:

$\left(\begin{matrix} cos(\frac{2\pi}{3}) & sin(\frac{2\pi}{3}) \\-sin(\frac{2\pi}{3}) & cos(\frac{2\pi}{3})\end{matrix}\right)$

Is this correct?

Thanks.

**Deveno** · November 20th 2011, 09:50 PM

normally an element x of $\mathbb{R}^2$ is written as (x,y) (sometimes as $(x_1,x_2)$ ). so for the example here:

$T(x,y) = (\cos(2\pi/3)x+\sin(2\pi/3)y, \cos(2\pi/3)y-\sin(2\pi/3)x)$

you can check your answer by multiplying your two matrices together, and verifying you get the 2x2 identity matrix.

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