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    Hi there im trying to understand this example in this textbook on matrix transformations:

    Describe the geometrical effect of the transformation given algebraically by:

    x' = 2x
    y' = 2y

    by applying a suitable matrix to the unit square and illustrating the effect graphically

    It starts the working out with a matrix

    [2 0
    0 2 ]

    Where did they get that matrix from? And in previous examples theyve used the
    [1 0
    0 1]

    which i thought by definition was the identity matrix and isnt used for anything.

    Please help confused!
    Thanks

    :confused:
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    (Original post by marek35)
    It starts the working out with a matrix

    [2 0
    0 2 ]

    Where did they get that matrix from?
    Well if the "new" x-coordinate is twice the old x-coordinate, and the "new" y-coordinate is twice the old y-coordinate, then that would be the matrix.

    In general, if x' = ax+by and y'=px+qy then the matrix representing this transformation is \begin{pmatrix} a & p \\ b & q \end{pmatrix}, which you can check by looking at what happens to the points (1,0) and (0,1) under the transformation.

    (Original post by marek35)
    And in previous examples theyve used the
    [1 0
    0 1]

    which i thought by definition was the identity matrix and isnt used for anything.
    It is the identity matrix, but I'm not sure what you mean by "isn't used for anything"... I mean, it doesn't have any effect, but it's certainly used for a lot of things :p: Can you give me an example of where they've used this, though? I mean, the linear formation it represents is quite limited in scope...
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    (Original post by nuodai)
    Well if the "new" x-coordinate is twice the old x-coordinate, and the "new" y-coordinate is twice the old y-coordinate, then that would be the matrix.

    In general, if x' = ax+by and y'=px+qy then the matrix representing this transformation is \begin{pmatrix} a & p \\ b & q \end{pmatrix}, which you can check by looking at what happens to the points (1,0) and (0,1) under the transformation.

    It is the identity matrix, but I'm not sure what you mean by "isn't used for anything"... I mean, it doesn't have any effect, but it's certainly used for a lot of things :p: Can you give me an example of where they've used this, though? I mean, the linear formation it represents is quite limited in scope...
    ohhhhh I think ive got it

    So if we multiplied this matrix back out we'd get back to what we started with: x' = 2x and y'= 2y is that the reason for this transformation matrix?

    And I get that if we test this out on a shape by multiplying each point by that transformation matrix we'd be able to see its a stretch of SF 2 with centre origin.

    Thanks
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    Another question if it says for example 'find the image of (1,0)' what exactly are we looking for in simple terms?

    Thanks
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    (Original post by marek35)
    ohhhhh I think ive got it

    So if we multiplied this matrix back out we'd get back to what we started with: x' = 2x and y'= 2y is that the reason for this transformation matrix?

    And I get that if we test this out on a shape by multiplying each point by that transformation matrix we'd be able to see its a stretch of SF 2 with centre origin.

    Thanks
    Yup. Basically, if you have a matrix \begin{pmatrix} a & b \\ p & q \end{pmatrix} then you can write out the corresponding equations as:
    \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a &  b \\ p & q \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax+by \\ px+qy \end{pmatrix}

    So when you're told, say, that x'=3x+y and y'=-y then you can just read off what the matrix is going to be.

    (Original post by marek35)
    Another question if it says for example 'find the image of (1,0)' what exactly are we looking for in simple terms?
    All that means is "find where (1,0) goes where you apply the matrix to it". So if your matrix were \begin{pmatrix} a & b \\ p & q \end{pmatrix} then the image of (1,0) is \begin{pmatrix} a & b \\ p & q \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ p \end{pmatrix}, and so on.
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    (Original post by nuodai)
    Yup. Basically, if you have a matrix \begin{pmatrix} a & b \\ p & q \end{pmatrix} then you can write out the corresponding equations as:
    \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a &  b \\ p & q \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax+by \\ px+qy \end{pmatrix}

    So when you're told, say, that x'=3x+y and y'=-y then you can just read off what the matrix is going to be.



    All that means is "find where (1,0) goes where you apply the matrix to it". So if your matrix were \begin{pmatrix} a & b \\ p & q \end{pmatrix} then the image of (1,0) is \begin{pmatrix} a & b \\ p & q \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ p \end{pmatrix}, and so on.
    Gotcha, thanks alot
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    Another question:

    A transformation is given by x' =x and y'=y-2x

    a) Find the image of the point P(3,0)

    So i went about it like this: x'= 3 and y' = 0-2(3) -> y' = -6 --> P'= (3,-6).

    b) Under this transformation the point S has image (4,-3) find the co-ordinates of point s.

    so S' = (4,-3) --> x=x' --> x = 4

    -3 = y-2(4) --> y = -5.

    --> S = (4,-5)

    Is this working out correct?

    The reason I ask is, this feels abit weird as its not in matrix form.
    Thanks
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    (Original post by marek35)

    -3 = y-2(4) --> y = -5.
    In general, yes, it's fine, apart from the bit in red - arithmetic slip!
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    (Original post by marek35)
    Another question:

    A transformation is given by x' =x and y'=y-2x

    a) Find the image of the point P(3,0)

    So i went about it like this: x'= 3 and y' = 0-2(3) -> y' = -6 --> P'= (3,-6).

    b) Under this transformation the point S has image (4,-3) find the co-ordinates of point s.

    so S' = (4,-3) --> x=x' --> x = 4

    -3 = y-2(4) --> y = -5.

    --> S = (4,-5)

    Is this working out correct?

    The reason I ask is, this feels abit weird as its not in matrix form.
    Thanks
    It seems fine, but I think the point of the question was that you were meant to find the matrix and use it for the first part, and then you were meant to find its inverse to use for the second part.
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    (Original post by ghostwalker)
    In general, yes, it's fine, apart from the bit in red - arithmetic slip!
    Oh yes, excuse my tiredness. Should be 5

    Cheers!
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    (Original post by nuodai)
    It seems fine, but I think the point of the question was that you were meant to find the matrix and use it for the first part, and then you were meant to find its inverse to use for the second part.
    Inverse? Dont think that is taught in FP1 on the AQA syllabus.

    Thanks though.
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    (Original post by marek35)
    Inverse? Dont think that is taught in FP1 on the AQA syllabus.

    Thanks though.
    I almost didn't believe you, but upon checking the syllabus it seems that matrix inverses don't come up until FP4... which is ridiculous, but let's go with that. In that case, you did the second part pretty much as you should have, although you could still have found the matrix for the first part.
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    (Original post by nuodai)
    I almost didn't believe you, but upon checking the syllabus it seems that matrix inverses don't come up until FP4... which is ridiculous, but let's go with that. In that case, you did the second part pretty much as you should have, although you could still have found the matrix for the first part.
    Yea I agree.

    How could I have done this question in a matrix?

    Thanks
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    (Original post by marek35)
    Yea I agree.

    How could I have done this question in a matrix?

    Thanks
    Well the first part is just finding an image, so you find what the matrix is from the equations and then just apply it to the vector \begin{pmatrix} 3 \\ 0 \end{pmatrix}.

    For the second part, as you know, you can find a matrix when you're told what x',y' are in terms of x,y. So to find its inverse you first find what x,y are in terms of x',y', and then find what matrix represents that transformation.

    In general, if \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} and ad-bc \ne 0 then \begin{pmatrix} x \\ y \end{pmatrix} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}. We call ad-bc the determinant of the matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} and we call \dfrac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} its inverse.
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    (Original post by nuodai)
    Well the first part is just finding an image, so you find what the matrix is from the equations and then just apply it to the vector \begin{pmatrix} 3 \\ 0 \end{pmatrix}.

    For the second part, as you know, you can find a matrix when you're told what x',y' are in terms of x,y. So to find its inverse you first find what x,y are in terms of x',y', and then find what matrix represents that transformation.

    In general, if \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} and ad-bc \ne 0 then \begin{pmatrix} x \\ y \end{pmatrix} = \dfrac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}. We call ad-bc the determinant of the matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} and we call \dfrac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} its inverse.

    AH yeah got the first bit

    ill do the 2nd bit tomorrow.



    Thanks for all your help!
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    I found another way of presenting part b) in a matrix.

    basically as we know that s'= (4,-3) I let 4= x' and -3=y'

    so that it becomes:

    4
    -3

    And we know that the matrix transformation that will be applied is:

    1 0
    -2 1 But we dont know what the original co-ordinates are so i let that =

    x
    y

    Multiplied it all out and solved and got S= (4,5).
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    So wanted to clarify a couple of things, if we have co-ordinates e.g. (4,5) that would be put in a matrix like this?

    4
    5

    so in general (x,y) = x
    x
    y

    and presenting algebraic stuff in a matrix e.g. x' = 2x-2y and y' = 3x+4y would be like this:

    x' = 2 -2 x
    y' = 3 4 .y

    and presenting x' = y and y'=x would be:

    x' = 0 1 x
    y' = 1 0 y
    so in general: x' = ax+by and y' = cx+dy then in a matrix it is:
    x' = a b x
    y' = c d y
    Just want to make sure I understand everything so far (hate this chapter).

    Thanks!
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    bump D:.
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    Hey again... this time its a question on rotations :P.

    How would I go about doing these two questions?

    Give a full description of the transformations given by the following matrices:


    I get that this is a rotation matrix about the origin but not sure how to work it out and that this matrix is used:


    For the first one I get that sin theta = 0 and cos theta = -1 and so tan theta = 0/-1 so theta = 0 which seems to be incorrect.

    And for the second one I did tan theta = -1 so theta= -45 which would be a clockwise rotation of 45 degree about the origin, is also incorrect. But if I try this on some other questions I get the right answer.

    Please help.
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    (Original post by marek35)
    Please help.
    For the first one, tan theta = 0 does not imply theta = 0, there is another possiblility.

    You need to be thinking of the values of sin, cos, tan, in the full four quadrants.
 
 
 
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