Further pure 1 matrices

I know it sounds stupid to make a discussion just to get the answer to one single question. But on this fp1 paper I was doing, I couldnt figure out how to answer part bii on the image I attached. I looked up the mark scheme and found the correct answer for it, but can someone help me on these types of matrices questions please ? Like how you are meant to know to take a number out of the elements of the matrix and make that the scale factor and how to know the right angle to use for a rotation(as there can be more than one angle) . Just basically a method to work out any qurstion like this without fault please, thanks anyone who can help :smile:

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Reply 1

Smaug123

Original post by JahJah

I know it sounds stupid to make a discussion just to get the answer to one single question.

No problem - it's the preferred way, because then all the discussion about it is collected in one place.

But on this fp1 paper I was doing…

Do you know that the image of the vector

(1,0)

is the first column of the matrix, and the image of

(0,1)

is the second column? In general, the matrix has columns {the image of the first basis vector, the image of the second basis vector, …}.

That means you can do it just by drawing a diagram and working out what kind of transformation converts the basis vectors (1,0),(0,1) to (-1, sqrt(3)), (-sqrt(3),-1).

A more mechanical way to do it: work out the determinant of A. That is the area (in 2d; in 3d it's volume, etc) scale factor of the transformation. In this case it's 4, so somewhere there's going to be a transformation that makes area bigger by a factor of 4. That makes length bigger by a factor of 2, so there's going to be a scale by 2 in there somewhere. Might as well carry it out last, so what we have is that A is "some transformation with matrix {{-1/2, -sqrt(3)/2)}, {-sqrt(3)/2, -1/2}}, followed by a scale by 2". Does the number sqrt(3)/2 ring any bells?

Reply 2

JahJah

Yes, sin of 60. I know the general identity for a matrix rotation(cosx-sinx)
(Sinx cosx)
So when I saw -sinroot3/2 was 60 I got confused so I used a cast diagram and found out that it is also 120 degrees, which made a bit more sense to the whole thing I guess.
you said about drawing a diagram to work it out, could you possibly draw one and send it to me ? Doing other rotations I used diagrams to help me. Such as I transforming to (cosx -sinx)
(Sinx cosx)
I use the unit circle to work out the angle of rotation.
Thanks for your help

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Reply 3

Smaug123

Original post by JahJah

This might help.

Hopefully it's legible.

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Reply 4

davros

Original post by JahJah

Original post by Smaug123

This might help.
Hopefully it's legible.

Looking at the order of the parts of the question, you could almost conjecture that A is a combination of an enlargement and a rotation. Since it's easy to show that A^3 = 8I (which is what they ask you to prove), you can see that the scale factor must be 2, and the angle of rotation for A alone must be 120 degrees, since if you cube A that is the same as rotating three times, and this must get you back to your original position.