# Linear Algebra question

1. Sep 30, 2006

### Perrry

Hi this is my first post here. So be nice now I´ve have a problem with a question:

$$T:\mathbb{R}^2\rightarrow\mathbb{R}^2$$ first reflects points through the vertical x2-axis and then rotates points $$\pi$$/2 radians.

/Perrry

2. Sep 30, 2006

### Daverz

What's the question?

3. Sep 30, 2006

What does 'x2-axis' mean? You mean like the y-axis in an orthogonal x-y coordinate system? What is your direction of rotation? Do you have to write the operator T in matrix form? State your question more clear, please.

4. Sep 30, 2006

### Perrry

Okay sorry my english isn´t that good. I missed some trival parts....

T is a linear tranformation. And i shall find the standard matrix of T. The x2 axis is the Y-axis. And the rotation should be counterclockwise.

//Perrry

5. Sep 30, 2006

To illustrate the problem, place a point T1 somewhere in the x-y coordinate system. Then follow the rules and get T1' and T1''. T1'' represents the point you get after the operator T is applied. So, write a matrix equation of form T1 T = T1'', whete T is your linear operator, which is a 2x2 matrix. Solve the equation, and you should get the elements of T.

6. Oct 1, 2006

### Perrry

Okay. The first matrix that reflects the points through the y-axis is $$\left[ \begin{array}{cc} -1 & 0\\ 0 & 1 \end{array} \right]$$

and then the matrix that rotate them counterclockwise is $$\left[ \begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right]$$

I should write them in a matrix that first reflects the points through the x-axis and then rotate them pi/2. Could i just write:

$$T= \left[ \begin{array}{cc} -1 & 0\\ 0 & 1 \end{array} \right]$$ $$\left[ \begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right]$$

Is the matrix finished there? And could i just put in the points (x,y) into the matrix?

Or could i write?:

$$T= \left[ \begin{array}{c} -x \\ y \end{array} \right]$$ $$\left[ \begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} \right]$$

And then put in the x,y if i had the values. Have patience guys I´m a newbie at this...

//Perrry

Last edited by a moderator: Oct 1, 2006