Line Image Transformation with 3x3 Matrix - Vector Solution

  • Thread starter Thread starter takercena
  • Start date Start date
  • Tags Tags
    Transformation
AI Thread Summary
The discussion focuses on finding the image of the line defined by r = i - j + n(i + j + k) under a specified 3x3 matrix transformation. The transformation is applied by multiplying the matrix by the vector representation of the line, leading to confusion about the parameter n and its role in the transformation. Clarification is provided that the image of a vector x is obtained by the matrix multiplication Mx, while the inverse matrix is used to find an inverse image x such that Mx = y. Participants express uncertainty about when to use the inverse matrix in the context of transformations. Understanding these concepts is essential for correctly applying matrix transformations in vector form.
takercena
Messages
25
Reaction score
0
Please help me solve this question:
Question: Find the images of the lines r = i - j + n(i + j + k) under transformation
M (matrix 3 x 3) =
(2 1 4)
(3 5 1)
(1 2 0)
in vector form.
 
Physics news on Phys.org
The image under the transformation is just that matrix times the vector? But what is n? My first thought was that it was the parameter but then that is just a single line. If that is correct then r= i- j+ n(i+ j+ k)= (n+1)i+ (n-1)j+ nk and its image is
\left[\begin{array}{ccc}2 & 1 & 4 \\ 3 & 5 & 1 \\ 1 & 2 & 0\end{array}\right]\left[\begin{array}{c} n+1 \\ n-1 \\ n\end{array}\right]
 
Thank you. There is some confusion here. I am confuse when to use inverse matrix to find transformation. Can you explain this. Example : 1. Find the images under the transformation
 
Last edited:
I have no idea what you mean by that. You use the inverse matrix in order to find an "inverse image": x such that Mx= y. Given a matrix M, the image of x is, by definition Mx.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top