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    Write out the 3x3 matrix that would be used to translate a point in the x-y plane by a distance a in the x direction and b in the y direction. Show this matrix is correct by applying it to an arbitrary point represented by the vector [x y 1].
    x direction = a, y direction = b.

    Whats my starting 3x3 matrix? What do I need to prove the matrix is correct? Multiply the 3x3 matrix by [x y 1] ?
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    (Original post by spikeymike)
    x direction = a, y direction = b.

    Whats my starting 3x3 matrix? What do I need to prove the matrix is correct? Multiply the 3x3 matrix by [x y 1] ?
    \begin{pmatrix} x_1 & x_2 & x_3 \\x_4 & x_5 & x_6 \\x_7 & x_8 & x_9 \end{pmatrix} \times  \begin{pmatrix} x \\y \\1 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    I think that's correct. I couldn't help but try helping despite being a bit crap with matrices
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    (Original post by Noble.)
    \begin{pmatrix} x_1 & x_2 & x_3 \\x_4 & x_5 & x_6 \\x_7 & x_8 & x_9 \end{pmatrix} \times  \begin{pmatrix} x \\y \\1 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    I think that's correct. I couldn't help but try helping despite being a bit crap with matrices
    Thanks for trying.

    I'll wait to see if anyone can confirm it. :holmes:
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    Yes, that's correct. (You still need to find what x_1, ..., x_9 need to be, of course).
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    (Original post by DFranklin)
    Yes, that's correct. (You still need to find what x_1, ..., x_9 need to be, of course).
    Of course


    (Original post by spikeymike)
    Thanks for trying.

    I'll wait to see if anyone can confirm it. :holmes:
    Should be relatively straightforward from there, just compare x, y and constant coefficients essentially.
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    (Original post by Noble.)
    \begin{pmatrix} x_1 & x_2 & x_3 \\x_4 & x_5 & x_6 \\x_7 & x_8 & x_9 \end{pmatrix} \times  \begin{pmatrix} x \\y \\1 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    I think that's correct. I couldn't help but try helping despite being a bit crap with matrices

    Ok so how does this:

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    prove that the "matrix is correct by applying it to an arbitrary point represented by the vector [x y 1]." ? :holmes:
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    (Original post by spikeymike)
    Ok so how does this:

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    prove that the "matrix is correct by applying it to an arbitrary point represented by the vector [x y 1]." ? :holmes:
    Well to me the question insinuates that you should already know the 3x3 matrix and you need to use it on [x, y, 1] to show that is does indeed produce [x+a, x+b, 1]. I've just reverse engineered the question and used those two sets of points to work out what the matrix should be. Since it seemed like you were trying to work out the matrix. I suppose once you've derived the matrix using those points, you'll have also shown that it's true since you've applied it to [x,y,1] at the same time.
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    Well, have you actually answered part (a) yet?
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    (Original post by DFranklin)
    Well, have you actually answered part (a) yet?
    I thought post #2 was the answer? :ashamed2:
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    (Original post by spikeymike)
    I thought post #2 was the answer? :ashamed2:
    No, you need to find the values of x_1, x_2, ..., x_9

    Just look at variable/constant coefficients.
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    (Original post by Noble.)
    No, you need to find the values of x_1, x_2, ..., x_9

    Just look at variable/constant coefficients.
    They're all 1? :dontknow:
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    (Original post by spikeymike)
    Ok so how does this:

    L.H.S. = \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix}

    prove that the "matrix is correct by applying it to an arbitrary point represented by the vector [x y 1]." ? :holmes:
    \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    So

    xx_1 + yx_2 + x_3 = x + a

    There's no y variable on the RHS, so there can't be on the left, what does this tell you x_2 must be?
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    (Original post by Noble.)
    \begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    So

    xx_1 + yx_2 + x_3 = x + a

    There's no y variable on the RHS, so there can't be on the left, what does this tell you x_2 must be?
    0?
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    (Original post by spikeymike)
    0?
    Yep

    Following the same logic, you should be able to see what x_1, x_3 are equal to. Then follow the rule for the next row, and the last row.
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    (Original post by noble.)
    yep :d

    following the same logic, you should be able to see what x_1, x_3 are equal to. Then follow the rule for the next row, and the last row.
    (1, 0, 0)
    (0, 1, 0)
    (0, 0, 1)

    :holmes:
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    Not quite.
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    (Original post by spikeymike)
    (1, 0, 0)
    (0, 1, 0)
    (0, 0, 1)

    :holmes:
    You've already worked out x_2 = 0

    xx_1 + y \times 0 + x_3 = x + a
    xx_1 + x_3 = x + a

    I'll give you a clue, both x_3 and a are constants.
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    (Original post by Noble.)
    You've already worked out x_2 = 0

    xx_1 + y \times 0 + x_3 = x + a
    xx_1 + x_3 = x + a

    I'll give you a clue, both x_3 and a are constants.
    (1, 0, a)
    (0, 1, b)
    (0, 0, 1)

    :gthumb:
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    (Original post by spikeymike)
    (1, 0, a)
    (0, 1, b)
    (0, 0, 1)

    :gthumb:
    Yep
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    So the final answer for part a is:

     \begin{pmatrix} x_1 & x_2 & x_3 \\x_4 & x_5 & x_6 \\x_7 & x_8 & x_9 \end{pmatrix} \times  \begin{pmatrix} x \\y \\1 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    =>\begin{pmatrix} xx_1 + yx_2 + x_3 \\xx_4 + yx_5 + x_6 \\xx_7 + yx_8  +x_9 \end{pmatrix} = \begin{pmatrix} x+a \\y+b \\1 \end{pmatrix}

    =\begin{pmatrix} 1, 0, a \\0, 1, b \\0, 0, 1 \end{pmatrix}

    :holmes:
 
 
 
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