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Description of matrices

How do you describe that transformations a matrix does?
eg;

T=[[br]1−2[br]21[br]]T = \begin{bmatrix}[br]1 & -2\\ [br] 2& 1[br]\end{bmatrix}


Give a full geometrical description of T.
(edited 8 years ago)
Original post by ubisoft
How do you describe that transformations a matrix does?
eg;

[1 -2]
[2 1]


Give a full geometrical description of T

Recognise the (+−++)\begin{pmatrix}+&-\\+&+\end{pmatrix} pattern in the signs. What standard transformation matrix do you know with this sign pattern? How can you modify it?
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Original post by morgan8002
Recognise the (+−++)\begin{pmatrix}+&-\\+&+\end{pmatrix} pattern in the signs. What standard transformation matrix do you know with this sign pattern? How can you modify it?


Yeah, I know it contains an anticlockwise rotation. How do you modify it to fit the standard form, so you can take inverse sin/cos? Is it just guesswork or is there a method?
Original post by ubisoft
Yeah, I know it contains an anticlockwise rotation. How do you modify it to fit the standard form, so you can take inverse sin/cos? Is it just guesswork or is there a method?

Good. You can't take inverse sin or cos because there may be(in this case is) a scale factor.

tan⁡θ=21=2\tan\theta = \frac{2}{1} = 2.
θ=63.4o\theta = 63.4^o (3sf)

We know that there isn't a real theta for which sin theta = 2, so there must also be a scale factor associated with the transformation.

Divide 2 by sin of the angle to get the scale factor.

Edit: if you know how to calculate the discriminant, you can spot scale factors more easily. If you don't, then don't worry.
(edited 8 years ago)
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Original post by morgan8002
Good. You can't take inverse sin or cos because there may be(in this case is) a scale factor.

tan⁡θ=21=2\tan\theta = \frac{2}{1} = 2.
θ=63.4o\theta = 63.4^o (3sf)

We know that there isn't a real theta for which sin theta = 2, so there must also be a scale factor associated with the transformation.

Divide 2 by sin of the angle to get the scale factor.

Edit: if you know how to calculate the discriminant, you can spot scale factors more easily. If you don't, then don't worry.


Sorry, where did you get tan from? Isn't the general form:

[[br]cosx−sinx[br]sinxcosx[br]]\begin{bmatrix}[br]cosx & -sinx\\ [br] sinx& cosx[br]\end{bmatrix}
Original post by ubisoft
Sorry, where did you get tan from? Isn't the general form:

[[br]cosx−sinx[br]sinxcosx[br]]\begin{bmatrix}[br]cosx & -sinx\\ [br] sinx& cosx[br]\end{bmatrix}

Yeah. Divide the 21(sin x) entry by the 11(cos x) entry to get tan x.
before we fond out the scale factor, the modified form is (kcos⁡x−ksin⁡xksin⁡xkcos⁡x)\begin{pmatrix} k\cos x & -k\sin x\\ k\sin x& k\cos x\end{pmatrix}
(edited 8 years ago)
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Original post by morgan8002
Yeah. Divide the 21(sin x) entry by the 11(cos x) entry to get tan x.
before we fond out the scale factor, the modified form is (kcos⁡x−ksin⁡xksin⁡xkcos⁡x)\begin{pmatrix} k\cos x & -k\sin x\\ k\sin x& k\cos x\end{pmatrix}

Did you just factorise out kcosx?

Does this work for all matrices or just this one?

edit nvm, I understand I think. Just equating and solving simultaneously?
(edited 8 years ago)
Original post by ubisoft
Did you just factorise out kcosx?

Does this work for all matrices or just this one?

edit nvm, I understand I think. Just equating and solving simultaneously?

There's a handful of ways to think about it. The easiest is to think of each element in the matrix as a number, so you have ksinx = 2 and kcosx = 1. If you divide ksinx by kcosx you just get tanx = 2.

This works for rotational matrices and the reflection in y=(tan⁡θ)xy= (\tan\theta)x ones(but that gives you tan⁡2θ\tan 2\theta).
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Original post by morgan8002
There's a handful of ways to think about it. The easiest is to think of each element in the matrix as a number, so you have ksinx = 2 and kcosx = 1. If you divide ksinx by kcosx you just get tanx = 2.

This works for rotational matrices and the reflection in y=(tan⁡θ)xy= (\tan\theta)x ones(but that gives you tan⁡2θ\tan 2\theta).


Ah yeah thanks. That's a nice method, the one my teacher told me depends more on guesswork, so I never really got it.

For tan2x, you can just take inverse tan and divide by 2? Or would you have to use the expansion formula for tan? (to get exact value in surd form)
Original post by ubisoft
Ah yeah thanks. That's a nice method, the one my teacher told me depends more on guesswork, so I never really got it.

For tan2x, you can just take inverse tan and divide by 2? Or would you have to use the expansion formula for tan? (to get exact value in surd form)


It's quicker to use the calculator, use arctan, divide by two and tan the answer. This might not give you the exact answer so you might need to round it to eg. (3sf). At A-level I think they'd be fine with this.

If you wanted to get the exact answer in surd form and the calculator didn't give it, use the double angle formula as you said. It will be tricky to find the exact answer though and takes some work.
(edited 8 years ago)
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Original post by morgan8002
It's quicker to use the calculator, use arctan, divide by two and tan the answer. This might not give you the exact answer so you might need to round it to eg. (3sf). At A-level I think they'd be fine with this.

If you wanted to get the exact answer in surd form and the calculator didn't give it, use the double angle formula as you said.


Yeah my exam board says to always give exact answers. Thanks for you help.
Original post by ubisoft
Yeah my exam board says to always give exact answers. Thanks for you help.

Is tan⁡2θ\tan 2\theta always rational in these questions? If not, then that's very mean unless they give a hint. There are STEP questions centred on this.
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Original post by ubisoft
Yeah my exam board says to always give exact answers. Thanks for you help.

Which exam board / module, out of interest?
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Original post by morgan8002
Is tan⁡2θ\tan 2\theta always rational in these questions? If not, then that's very mean unless they give a hint. There are STEP questions centred on this.


Yeah it's nothing too complicated. Most of the time it's in the form 1x\frac{1}{\sqrt{x}} so you can arrive at your answer by just playing around with the decimal. I just square my decimal and take the reciprocal and it's usually a whole number haha, bit of a cheat method though.
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Original post by notnek
Which exam board / module, out of interest?


This is from AQA FP4
Original post by ubisoft
Yeah it's nothing too complicated. Most of the time it's in the form 1x\frac{1}{\sqrt{x}} so you can arrive at your answer by just playing around with the decimal. I just square my decimal and take the reciprocal and it's usually a whole number haha, bit of a cheat method though.

Oh, ok. Double angle's fine then.

Original post by ubisoft
This is from AQA FP4

I'm doing this too. I hadn't seen any questions of this type but haven't done the more recent few past papers yet. Which paper is an example of this from?
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Original post by morgan8002
Oh, ok. Double angle's fine then.


I'm doing this too. I hadn't seen any questions of this type but haven't done the more recent few past papers yet. Which paper is an example of this from?


This was from Q8 Jan '09

http://www.gosford-hill.oxon.sch.uk/past-exam-papers/alevel-maths/alevel-maths/Further%20Pure/FP4/FP4%20Question%20Papers/AQA-MFP4-W-QP-JAN09.pdf
Original post by ubisoft


No I meant a question on the reflection in y=(tan⁡θ)xy=(\tan\theta)x.
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Original post by morgan8002
No I meant a question on the reflection in y=(tan⁡θ)xy=(\tan\theta)x.


Oh, I haven't seen that in fp4. I think that's only in fp1
Original post by ubisoft
Oh, I haven't seen that in fp4. I think that's only in fp1

Ok. I think they're allowed to ask it in FP4 but maybe they never have or they might have in a paper neither of us has done yet.

In FP1 they can't expect you to use double angle formulae so I think they only ask very simple ones eg. tan⁡2θ=1\tan 2\theta = 1.

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