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AS FM Matrices: can someone check this please?

New textbooks mean mistakes but I incorrectly identified a mistake earlier so can someone please double check this one for me?

The answer given is

\tan^{-1}(3)

\tan^{-1}(\frac{1}{3})

. *

Using a geometric approach I agree with

\tan^{-1}(3)

but I don't think

\tan^{-1}(\frac{1}{3})

is possible. There's a good chance that I'm being stupid so can someone please check? I'd be happy to post my working if someone agrees with the answers.

*

\pm \pi

was missed in the answers which is annoying.

Reply 1

Amullai

Wow i thought someone would have answered by now. I'm just trying it now. But you know more maths than me.

Reply 2

Amullai

Would tan^-1(1/3) give you the line y=x? That's the only explanation I can think of.

Reply 3

Amullai

Just did the mathsy stuff. It makes a line of invariant points so you're wrong I'm afraid... :frown:

I can show my work if you want me to.
It was just working out the angle in the answer and then working out the angle that turns y=2x to y=x It's about 18.4(+ or - n180 where n is an integer) degrees same as the answer.

Reply 4

Notnek

Original post by Amullai

It was just working out the angle in the answer and then working out the angle that turns y=2x to y=x It's about 18.4(+ or - n180 where n is an integer) degrees same as the answer.

I agree with this but S is a clockwise rotation matrix so this is wrong I think?

-\tan (\frac{1}{3})

would make sense to me. I can't post full explanation right now but I can later.

Reply 5

DFranklin

notnek

I believe the correct answers are:

\tan^{-1}(3)

maps the line y=x to the line x=-2y, which is perpendicular to the reflection line y = 2x. This line is invariant and points on it are reflected (multiplied by -1).

-\tan^{-1} \left(\frac{1}{3}\right)

maps the line y=x to the line y=2x which is the reflection line. This line is invariant and points on it are not changed at all.

(This is with clockwise rotations).