Matrices

If you didnt remember the matrices for relection in y=-x
How could you work it out?
How do i do cii)

Draw the unit vectors (1,0) and (0,1) on a sketch of a coordinate axes. Reflect both vectors in the line y=-x. The result will give you:

(1,0) -> (a,b)
(0,1) -> (c,d)

Then the matrix with columns (a,b) and (c,d) is the transformation matrix.

Draw the unit vectors (1,0) and (0,1) on a sketch of a coordinate axes. Reflect both vectors in the line y=-x. The result will give you:

(1,0) -> (a,b)
(0,1) -> (c,d)

Then the matrix with columns (a,b) and (c,d) is the transformation matrix.

Draw the unit vectors (1,0) and (0,1) on a sketch of a coordinate axes. Reflect both vectors in the line y=-x. The result will give you:

(1,0) -> (a,b)
(0,1) -> (c,d)

Then the matrix with columns (a,b) and (c,d) is the transformation matrix.

Like this?




Is this correct?

For coo o tried this?
But it's wrong..
How would I do these type of questions?

Thanks

Which part do you need help with : c)i or c)ii ?

cii) please

How can you work out the scale factor of the enlargement using the matrix?

http://www.thestudentroom.co.uk/attachment.php?attachmentid=411215&d=1432835197
I tried this to find h, the scale factor.

You can find the scale factor just by looking at the determinant of the matrix.

The determinant of the matrix gives you the area scale factor of the transformation. So what's the scale factor of the enlargement?

Ive never learned about determinants, but i know how to calculatre them?
How are they useful? What can you do with the determinant here?

Sorry, I think this is an AQA question in which case determinants are not part of your syllabus I believe.


Instead you can use part c)i to help you: $A^2=12I$

$A^2$ represents a transformation of A followed by another one of A. The fact above shows that this is represented by the matrix

$12I = \begin{pmatrix}12 & 0 \\ 0 &12\end{pmatrix}$

Which is only an enlargement scale factor 12. Can you see how this shows that $A$ represents a reflection? And how can you use the above to find the enlargement scale factor of a single transformation using A?

You could have also drawn a sketch of this transformation using the unit vectors. Then it isn't too hard to find the scale factor of the enlargement using pythagoras.

I'm sorry I still do not understand how I could use this to find the rotation angle and enlargement.
Please could you just explain thoroughly how I would use it?
My exam is in a couple of days time, and my mine is somewhere else in other subjects atm.
I appreciate you trying to break it down for me but I realise it's taking me longer and longer to decipher what you have written.
Thanks

Do you understand that $A^2$ represents a transformation of $A$ followed by $A$?

So from part i, $\begin{pmatrix}12 & 0 \\ 0 &12\end{pmatrix}$ represents that combined transformation.


If $A$ represents a reflection and an enlargement then the above is telling you that if you do the reflection twice and the enlargement twice then you will end up with only an enlargement scale factor 12.

This makes sense because if you do a reflection twice then you will get back to where you started.


So just focussing on the enlargement, if you do the enlargement twice you end up with an enlargement scale factor 12. That means that a single enlargement must have scale factor $\sqrt{12}$ because $\sqrt{12}\times \sqrt{12} = 12$

i.e. if you enlarge with SF $\sqrt{12}$ twice then the combined enlargement has SF $12$.


If any of this doesn't make sense, please tell me the exact part where you got stuck with my explanation.

Ok I understand thanks.
So is the enlargement = root 12 since enlarging twice gives 12.
If the enlargement is root 12 then I take a factor of root 12 out of A? Then work out angle?

Yes.

But then sin 2x = -0.5
Which x = -15 degrees
Subbing in -15 degrees into coz 2x is not equal to -3/2root3 which is what I get when I take a factor out of A?

-15 is not the only solution of $\sin 2x=-0.5$

You need to find a solution of the two equations $\sin 2x = -0.5$ and $\cos 2x = -\frac{\sqrt{3}}{2}$.

Sorry, I've been busy recently and haven't been able to reply much.

Understood, thanks for your help!
But is this determinant way much quicker?