- #1

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:R

^{2}=> R

^{3}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:

[tex]\left[ \begin{array}{ccc} 0 & -2 & 0 \\ 6 & -3 & 0 \\ 5 & 1 & 0 \end{array} \right]

[/tex] * [x, y, 1]^T = [tex]

\left[ \begin{array}{ccc} -2y\\6x-3y\\5x+y \end{array} \right]

[/tex]

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:

[tex]

\left[ \begin{array}{ccc} 1&0&0\\0&1/2&-sqrt(3)/2 \\ 0&sqrt(3)/2&1/2 \end{array} \right]

[/tex]

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!