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Right question, I am stuck. We have been working on matrices and I think I understand them, however I have no idea how to apply these to this transformation question.

Consider the points O = (0; 0; 0), A = (1; 3; 3) and B = (-2; -6; 4) together with the transformation which carries out a scaling to increase distances in the x direction by a factor of two, decrease those in the y - direction by a factor of three and leave distances in the z - direction unchanged. It also leaves the point A unchanged. What is the matrix associated with this transformation? Use the matrix to find the images of the following

(a) the point O
(b) the point A
(c) the point B
(d) the vector A->B
(e) the vector O->B
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2 Answers 2

up vote 3 down vote accepted

If the matrix you're looking for is a 3x3 matrix, then no such matrix exists, since

$A = x + 3y + z$, so $MA = Mx + 3My + 3Mz = (2,1,3) \neq A$.

But since you make the distinction between points and vectors, are you maybe working in projective space?

Then take the matrix $ N = \begin{bmatrix} 1 & 0 & 0 & A_x \\ 0 & 1 & 0 & A_y \\ 0 & 0 & 1 & A_z \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $

and let

$M = NSN^{-1}$,

where $S$ is the diagonal matrix with entries $(2,\frac{1}{3},1,1)$.

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The above mentioned transformation can be written as an affine mapping in $R^3$: $$ \begin{bmatrix} x' \\ y' \\ z' \\ \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1/3 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} + \begin{bmatrix} -1 \\ 2 \\ 0 \\ \end{bmatrix} $$ It can also be written as a linear mapping in $R^4$ by introducing homogenous coordinates:

$$ \begin{bmatrix} x' \\ y' \\ z' \\ w' \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 & -1 \\ 0 & 1/3 & 0 & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} $$

Using any one of these equations we get for the images (...)' of (...):

(a) O' = (-1, 2, 0)
(b) A' = (1, 3, 3) = A
(c) B' = (-5, 0, 4)
(d) (A -> B)' = B' - A' = (-6, -3, 1)
(e) (O -> B)' = B' - O' = (-4, -2, 4) 
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