Line Image Transformation with 3x3 Matrix - Vector Solution

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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.
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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.
 
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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
 
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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 tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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