Matrix Transformation: Linear Algebra Final Question Explained

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SUMMARY

The discussion focuses on a linear transformation S: R² → R³ defined by S[x, y]ᵀ = [-2y, 6x - 3y, 5x + y]ᵀ. The associated matrix A for this transformation is identified as a 3x2 matrix, specifically [[0, -2], [6, -3], [5, 1]]. Additionally, the transformation T involves a rotation through 60 degrees about the x-axis followed by a reflection in the x-y plane, leading to a separate matrix B. The volume of the image of a unit cube under transformation T is determined to be 1, indicating that rotation does not alter volume.

PREREQUISITES
  • Understanding of linear transformations in linear algebra
  • Familiarity with matrix representation of transformations
  • Knowledge of rotation and reflection matrices in 3D space
  • Concept of volume transformation under linear mappings
NEXT STEPS
  • Study the derivation of the standard matrix for linear transformations in R² and R³
  • Learn about rotation matrices, specifically for 3D transformations
  • Research the properties of volume preservation under linear transformations
  • Explore the implications of mapping dimensions in linear algebra
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Students of linear algebra, particularly those preparing for exams or needing clarification on linear transformations, matrix representations, and volume calculations in higher dimensions.

Melawrghk
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Homework Statement


Hi everyone!

I'm trying to figure out this question that was on my Linear Alg final last year with the intent of maybe appealing the mark. I'd really appreciate if you could explain this to me.

Question: Let S:R2 => R3 be a linear transformation given by S[x, y]T = [-2y, 6x-3y, 5x+y]T

a) Find the associated matrix A of T
b) Find the matrix B of the 3-dimensional transformation T composed of
i) a rotation through 60 degrees about the x axis, followed by
ii) a reflection in the x-y plane
c) If C is a cube of unit volume, what is the volume of the image T(C)


Homework Equations


None


The Attempt at a Solution


First of all, I don't get why they switched from S to T all the sudden. For part (a) , I had:
\left[ \begin{array}{ccc} 0 &amp; -2 &amp; 0 \\ 6 &amp; -3 &amp; 0 \\ 5 &amp; 1 &amp; 0 \end{array} \right]<br /> * [x, y, 1]^T = <br /> <br /> \left[ \begin{array}{ccc} -2y\\6x-3y\\5x+y \end{array} \right]<br /> <br />
The first matrix is A that they want.
b) So for this part I got confused, I wasn't sure if they were talking about the same T (in that case, how can it have two corresponding matrices?). I assumed it was a new one and thus I got:
<br /> <br /> \left[ \begin{array}{ccc} 1&amp;0&amp;0\\0&amp;1/2&amp;-sqrt(3)/2 \\ 0&amp;sqrt(3)/2&amp;1/2 \end{array} \right]<br /> <br />
I got those values by drawing a y-z plane and "rotating" coordinates of two unit vectors that correspond to y&z coordinates, because the x-coordinate shouldn't change.
c) For this part I got even more confused because now I had no idea which T they wanted, the one from part (a) or (b). I did work for both - I did triple product in both cases.
For T from part (a) I got V=0 because of that last column being all zeros. And for T from part (b) I got V=1, which makes sense because it's just a rotation and shouldn't change the volume.

Phew, that's it. Can someone offer me some feedback on this? I really want to get my A back in that class, but I'm not sure I have a case... Thanks in advance!
 
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Your answer for the standard matrix A isn't correct. Note that it is a mapping from R2 to R3, so from this you should deduce that the required matrix is of size 3x2. And there's a difference between R2 and R3; you can't just add a 0 to the third dimension if nothing is specified.
 

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