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    Part a I found fine thanks to earlier help but part b is confusing.

    I tried to use determinant to guide me but to no avail.

    How do I do this? It's only worth one mark so I feel like I'm missing something easy.
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    (Original post by Ravster)
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    Part a I found fine thanks to earlier help but part b is confusing.

    I tried to use determinant to guide me but to no avail.

    How do I do this? It's only worth one mark so I feel like I'm missing something easy.
    If you multiply a 2x2 matrix by the 2x2 identity matrix you end up with the same matrix.

    As A^n is just doing the transformation n times, what it's really asking you is how many times do you need to carry out the transformation to get back to where you started.
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    (Original post by Ravster)
    Name:  Screenshot_20160513-234201.jpg
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    Part a I found fine thanks to earlier help but part b is confusing.

    I tried to use determinant to guide me but to no avail.

    How do I do this? It's only worth one mark so I feel like I'm missing something easy.
    If you have some coordinates \begin{pmatrix}x \\ y \end{pmatrix} and you multiply it by the identity matrix I, you will get the same thing: \begin{pmatrix}x \\ y \end{pmatrix}.

    As A is the rotation about the origin by certain angle, how many times will it have rotate by that angle to get to its original position k360^{\circ}

    Edit: Zacken has pointed out that you would have to get to multiple of 360^{\circ} which would be the LCM of the angle you found in part(a) and 360.
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    (Original post by Ravster)
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    Part a I found fine thanks to earlier help but part b is confusing.

    I tried to use determinant to guide me but to no avail.

    How do I do this? It's only worth one mark so I feel like I'm missing something easy.
    Think about what the angle of rotation A represents and think about how many multiples of that rotation you need to get to a multiple of 360 degrees, which will be the identity matrix.

    Remember that if A represents a rotation of \alpha then A^2 is a rotation of \alpha followed by a rotation of \alpha for a total rotation of 2\alpha. In general here A^n is a rotation of n\alpha degrees. So you need to find a n s,t n\alpha is a multiple of 360.
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    (Original post by shivtek)
    Part was 135 I think. Or 45. I can't remember. I is the identity as you know so what it's asking is how many times does one apply A to get back to the start. So you do 360/(135 or 45). I got 8/3 I think
    (Original post by Kvothe the arcane)
    If you have some coordinates \begin{pmatrix}x \\ y \end{pmatrix} and you multiply it by the identity matrix I, you will get the same thing: \begin{pmatrix}x \\ y \end{pmatrix}.

    A represents a transformation and it's saying that \mathrm{\underbrace{A \times A \times \times A \times .... \times A}_{n \ times}=X}

    As A is the rotation about the origin by certain angle, how many times will it have rotate by that angle to get to its original position 360^{\circ}
    I'm going to have to butt in here and say that you want a natural n; so n=\frac{8}{3} is incorrect. You don't want to rotate to get to 360^{\circ} but rather, you want to rotate to get to a natural multiple of 360^{\circ}. Which is an important point here.
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    (Original post by Zacken)
    ...
    That makes sense. You can't have a fraction of a multiplication. It would have made sense to evaluate a.
    Thanks for pointing it out.
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    (Original post by Zacken)
    I'm going to have to butt in here and say that you want a natural n; so n=\frac{8}{3} is incorrect. You don't want to rotate to get to 360^{\circ} but rather, you want to rotate to get to a natural multiple of 360^{\circ}. Which is an important point here.
    Away good point I didn't read "positive integer"
    So really, raising it to the power of 8 would work, as it would be 0 rotation modulo 360. Thanks for pointing that out!
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    (Original post by shivtek)
    Away good point I didn't read "positive integer"
    So really, raising it to the power of 8 would work, as it would be 0 rotation modulo 360. Thanks for pointing that out!
    Yep, 8 is the only correct answer here since it specifies n being the smallest positive integer satisfying the condition. Right on with the rotation 0 modulo 360!
 
 
 
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