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    Hi all,

    In the two attachments, you can see that the mark scheme shows that the enlargement is sqrt(12). It also shows how you pull this out as a linear enlargement and hence the original matrix is changed. I understand this.

    I was wondering how I should realise that I should take sqrt(12) from the original matrix? I understand how to do so, but I don't understand how I should have the inspiration to extract sqrt(12) from the matrix, especially as it is not a 'nice' number? For example, I tried taking out 2 and 3 from the matrix, but to no avail.

    Thank you!
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    (Original post by londoncricket)
    Hi all,

    In the two attachments, you can see that the mark scheme shows that the enlargement is sqrt(12). It also shows how you pull this out as a linear enlargement and hence the original matrix is changed. I understand this.

    I was wondering how I should realise that I should take sqrt(12) from the original matrix? I understand how to do so, but I don't understand how I should have the inspiration to extract sqrt(12) from the matrix, especially as it is not a 'nice' number? For example, I tried taking out 2 and 3 from the matrix, but to no avail.

    Thank you!
    Doesn't have the be sqrt(12), could have pulled out 2sqrt(3). Basically any number that makes the left over matrix one that you can easily see is a rotation matrix with a recognisable angle.
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    (Original post by londoncricket)
    Hi all,

    In the two attachments, you can see that the mark scheme shows that the enlargement is sqrt(12). It also shows how you pull this out as a linear enlargement and hence the original matrix is changed. I understand this.

    I was wondering how I should realise that I should take sqrt(12) from the original matrix? I understand how to do so, but I don't understand how I should have the inspiration to extract sqrt(12) from the matrix, especially as it is not a 'nice' number? For example, I tried taking out 2 and 3 from the matrix, but to no avail.

    Thank you!
    Consider:

    What do you know about the form of a reflection matrix?

    And what's the standard identity for sines and cosines?

    Hence the scale factor must be....
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    (Original post by Zacken)
    Doesn't have the be sqrt(12), could have pulled out 2sqrt(3). Basically any number that makes the left over matrix one that you can easily see is a rotation matrix with a recognisable angle.
    I wouldn't have had the thought to take out two numbers though


    (Original post by ghostwalker)
    Consider:

    What do you know about the form of a reflection matrix?

    And what's the standard identity for sines and cosines?

    Hence the scale factor must be....
    Uh, the form of a reflection matrix is a two by two matrix with sin(2theta) in two corners and cos(2theta) in the other two with a minus sign in front of the bottom cos(2theta).

    I don't know what the standard identity means?

    I am not sure as to how that would lead me to sqrt(12)?

    Thank you for both of your replies though, I appreciate it!
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    (Original post by londoncricket)

    I am not sure as to how that would lead me to sqrt(12)?
    I think it's better thought of as 2sqrt(3) instead of sqrt(12).
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    (Original post by londoncricket)
    I
    Uh, the form of a reflection matrix is a two by two matrix with sin(2theta) in two corners and cos(2theta) in the other two with a minus sign in front of the bottom cos(2theta).

    I don't know what the standard identity means?
    Cos^2 x+sin^2 x=1

    Apply that idea to your first row and you get 12, instead of 1, and since you're squaring each term you need to scale by root(12)
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    IF you take out a factor, say  h you get  \displaystyle \mathbf{A} =h\begin{bmatrix} -\frac{3}{h} & -\frac{\sqrt 3}{h} \\ -\frac{\sqrt 3}{h} & \frac{3}{h} \end{bmatrix} .
    As you can see that this is in the form of a reflection matrix (opposite sign of left diagonal elements).
     \cos 2\theta =-\frac{3}{h} and  \sin 2\theta = -\frac{\sqrt 3}{h} .
    If you draw a right angle triangle you should be able to see that the hypotenuse (by Pythagoras) is  \sqrt{12}=2\sqrt{3} . So  h=2\sqrt{3} , so now you know the scale factor of the enlargement which is  2\sqrt{3} .
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    (Original post by ghostwalker)
    Cos^2 x+sin^2 x=1

    Apply that idea to your first row and you get 12, instead of 1, and since you're squaring each term you need to scale by root(12)
    (Original post by B_9710)
    IF you take out a factor, say  h you get  \displaystyle \mathbf{A} =h\begin{bmatrix} -\frac{3}{h} & -\frac{\sqrt 3}{h} \\ -\frac{\sqrt 3}{h} & \frac{3}{h} \end{bmatrix} .As you can see that this is in the form of a reflection matrix (opposite sign of left diagonal elements). \cos 2\theta =-\frac{3}{h} and  \sin 2\theta = -\frac{\sqrt 3}{h} .If you draw a right angle triangle you should be able to see that the hypotenuse (by Pythagoras) is  \sqrt{12}=2\sqrt{3} . So  h=2\sqrt{3} , so now you know the scale factor of the enlargement which is  2\sqrt{3} .
    Thank you for this! Really useful! Teaming your two responses, I think that in the future I will take out 'h', as an enlargement factor and this can lead to (-3/h)^2 + (-sqrt(3)/h)^2 = 1, which solves for h = sqrt(12).

    Thank you both very much!
 
 
 
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