Matrix Transformation: Linear Algebra Final Question Explained

AI Thread Summary
The discussion revolves around a Linear Algebra final question involving a linear transformation S from R2 to R3. The participant seeks clarification on finding the associated matrix A, the composition of transformations, and the volume of the image of a unit cube under the transformation. For part (a), the participant incorrectly identified matrix A as 3x3 instead of the correct 3x2 size, which is crucial since the transformation maps from R2 to R3. In part (b), confusion arises regarding the transformation T and its associated matrices, while part (c) leads to uncertainty about which transformation to use for volume calculations. Overall, the participant is looking for feedback to support their appeal for a higher grade.
Melawrghk
Messages
140
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
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...

Similar threads

Replies
3
Views
2K
Replies
69
Views
8K
Replies
13
Views
2K
Replies
11
Views
2K
Replies
8
Views
1K
Replies
8
Views
2K
Replies
31
Views
2K
Replies
5
Views
2K
Back
Top