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    Two matrices A and B are given by

     A= \frac{1}{3} <font color="#444444"><span style="font-family: verdana">\begin{pmatrix} 2 & -2 & -1 \\1 & 2 & -2 \\2 & 1 & 2 \end{pmatrix}

     B=\frac{1}{3}<font color="#444444"><span style="font-family: verdana">\begin{pmatrix} 1 & 2 & -2 \\2 & 1 & 2 \\-2 & 2 & 1 \end{pmatrix}.

    One is a reflection and the other a rotation. Suppose X is the rotation matrix and Y is the reflection matrix, determine which of A and B is the rotation and which is the reflection.

    I honestly have no idea how to start this problem. Any initial shove in the right direction would be appreciated.
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    (Original post by Music99)
    Two matrices A and B are given by

     A= \frac{1}{3} <font color="#444444"><span style="font-family: verdana">\begin{pmatrix} 2 & -2 & -1 \\1 & 2 & -2 \\2 & 1 & 2 \end{pmatrix}

     B=\frac{1}{3}<font color="#444444"><span style="font-family: verdana">\begin{pmatrix} 1 & 2 & -2 \\2 & 1 & 2 \\-2 & 2 & 1 \end{pmatrix}.

    One is a reflection and the other a rotation. Suppose X is the rotation matrix and Y is the reflection matrix, determine which of M and N is the rotation and which is the reflection.

    I honestly have no idea how to start this problem. Any initial shove in the right direction would be appreciated.
    Not sure what X, Y, M and N are meant to be, but one way to approach this would be to think about determinants.
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    (Original post by Mark13)
    Not sure what X, Y, M and N are meant to be, but one way to approach this would be to think about determinants.
    Opps wrote out a wrong bit see the edit. . I think X and Y are just meant to be a general rotation and reflection in 3 dimensions.
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    Well a pure rotation matrix in 3D has an eigenvalue of 1, amongst other things, yet neither of A, nor B does.

    Unless I made another slip!

    Edit: Made a slip. First has an eigenvalue of 1. :sigh:
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    (Original post by ghostwalker)
    Well a pure rotation matrix in 3D has an eigenvalue of 1, amongst other things, yet neither of A, nor B does.

    Unless I made another slip!
    We haven't covered eigenvalues yet. I'm not too sure all the question says is Suppose X is the rotation matrix and Y is the reflection matrix. Determine which of A and B is the rotation and which is the reflection.
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    (Original post by Music99)
    We haven't covered eigenvalues yet. I'm not too sure all the question says is Suppose X is the rotation matrix and Y is the reflection matrix. Determine which of A and B is the rotation and which is the reflection.
    Have you posted the question exactly as stated? there doesn't seem to be any need for the clause beginning "Suppose X..." - surely it could just ask which of A and B is the rotation and which is the reflection???
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    (Original post by davros)
    Have you posted the question exactly as stated? there doesn't seem to be any need for the clause beginning "Suppose X..." - surely it could just ask which of A and B is the rotation and which is the reflection???
    That's what I thought, maybe it's missing a bit of info? The exact wording is

    The matrices A and B are given by:

    And gives the matrices I gave above. One is a reflection and the other a rotation. Suppose X is the rotation matrix and Y is the reflection matrix. Determine which if A and B is the rotation and which is the reflection.
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    Bump.
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    (Original post by Music99)
    Bump.
    Could you do a scan of the page? It still seems confusing with the X, Y.
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    (Original post by ghostwalker)
    Could you do a scan of the page? It still seems confusing with the X, Y.
    Yeah sure, I just realized I used A and B for the matrices instead of M and N. I think maybe not all the question printed or something
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    (Original post by Music99)
    Yeah sure, I just realized I used A and B for the matrices instead of M and N. I think maybe not all the question printed or something
    I think you can try, if a matrix represents a reflection, then squaring it should give you the identity matrix.
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    (Original post by ghostwalker)
    I think you can try. If a matrix represents a reflection, then squaring it should give you the identity matrix.
    I'll give that a try, the X and Y thing is just weird.
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    (Original post by Music99)
    I'll give that a try, the X and Y thing is just weird.
    Sorry, just realised that seemed a bit arrogant. I put in a full stop thoughtlessly.
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    (Original post by ghostwalker)
    Sorry, just realised that seemed a bit arrogant. I put in a full stop thoughtlessly.
    It's cool. Okay so I did that and got the 2nd matrix as the reflection, how did you know that squaring it like that would work?
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    Oh I thought this was a question about the film The Matrix


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    (Original post by holson)
    Oh I thought this was a question about the film The Matrix
    It is not "the one", Zathrus. (To mix my universes)
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    (Original post by holson)
    Oh I thought this was a question about the film The Matrix


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    XD! If only. That would be epic, Neo you must work out the optimal punching angle to hit agent smith in the face .
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    (Original post by Music99)
    It's cool. Okay so I did that and got the 2nd matrix as the reflection, how did you know that squaring it like that would work?
    A reflection of a reflection takes you back to where you started. Hence....

    It's not a definitive test as repeating a rotation of 180 degrees would do the same thing.

    But if you know one isn't a reflection and the square of one isn't the identity matrix, then that's the one that isn't a reflection.
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    (Original post by ghostwalker)
    A reflection of a reflection takes you back to where you started. Hence....

    It's not a definitive test as repeating a rotation of 180 degrees would do the same thing.

    But if you know one isn't a reflection and the square of one isn't the identity matrix, then that's the one that isn't a reflection.
    Ah right that makes sense thank you!. If I get stuck on other parts is it okay if I quote you?
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    (Original post by Music99)
    If I get stuck on other parts is it okay if I quote you?
    Sure. Can't guarantee I'll be able to help though.
 
 
 
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