# Thread: Finding the 2 x 2 matrix

1. ## Finding the 2 x 2 matrix

Find the 2 by 2 matrix representing the transformation which maps triangle A onto triangle B.

I need, not just the solution but the method to it, pretty quickly. Appreciate any help you offer me, thanks!

As far as I can tell, it's a anti-clockwise 90 degree rotation. So would I work it out by...

1. Select the points (1,0) and (0,1) on the graph, and then the matrix is
(1 0
(0 1)
2. Find the images of these points under rotation
3. Write the position vectors of these images.
4. The position vector of the image of the first point becomes the first column.
5. The position vector of the image of the second point is the second column of the required matrix.

(1 0)
(0-1) ?

2. I was always taught a general transformation for a rotation:

$\left(\begin{array}{cc}Cos\theta&-Sin\theta\\Sin\theta&Cos\theta\end{array}\right)$

Where $\theta$ is the number of degrees anticlockwise that the shape is being rotated. I don't understand your method, admittedly, because I never had to use it, but I would suspect that there is a mistake within your working because it contradicts with my method, which is a general formula. I'll check your method more carefully and see if I can work out where the error lies - it may be that for whatever reason my formula doesn't apply here.

3. Not quite the matrix you give is a reflection across the x axis.

Notice that the angle in triangle A across from the hypotenuse is represented by

$2e_1+e_2$

This gets mapped to

$T(2e_1+e+2)=2T(e_1)+T(e_2)=-e_1+2e_2$

Picking another point on the triangle we get

$T(3e_1+e+2)=3T(e_1)+T(e_2)=-e_1+3e_2$

Now if we subtract the 1st equation from the 2nd we get

$T(e_1)=e_2$

This gives

$T(e_2)=-e_1$

$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

4. Aha! If you take the point (1,0), then the image under transformation becomes (0,1). So the first column is (0,1)

If you take the point (0,1) and rotate it 90 degrees, then it becomes (-1,0) which is the second column.

So this yields $\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$ which agrees with my formula.

Edit: was beaten by TES.

5. Ooookay, right! So the first column is (0,1), then the 2nd is (1,0), you rotate them anticlockwise 90degrees, then it becomes
(0 -1)
(1 0)

Got it, thanks Quacky! By the way, I don't get your joke about Descartes... He's the one who made the Cartesian system right?

6. Originally Posted by yorkey
Ooookay, right! So the first column is (0,1), then the 2nd is (1,0), you rotate them anticlockwise 90degrees, then it becomes
(0 -1)
(1 0)

Got it, thanks Quacky! By the way, I don't get your joke about Descartes... He's the one who made the Cartesian system right?
He also came up with the famous philosophical statement "I think, therefore I am".