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Math Help - Linear transformation, volume, problem

  1. #1
    MHF Contributor arbolis's Avatar
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    Linear transformation, volume, problem

    True or false :
    Let T:\mathbb{R}^3 \to \mathbb{R}^3 be a linear transformation defined by T(1,0,0)=(1,1,0), T(0,1,0)=(-2,1,1), T(0,0,1)=(0,3,-2).
    If C is the cube formed by the vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1), then T(C)=3^2.
    My attempt : By intuition its false because 3^2 seems more like an area, but it could be a volume.
    Ok, so I have formed [T]_c^{c} in order to find out that T(x,y,z)=(x-2y,x+y+3z,y-2z). Then I found the vertices passed by the transformation to be (0,0,0), (1,1,0), (-2,1,1) and (0,3,-2). Now it remains to calculate the volume of the solid formed by these vertices. I'm not really good when it comes to draw 3 dimensional solids. Especially this "prism" looking volume. I've no idea how to find out its volume.


    Oh.... let me see, I remember I've read in a calculus III book that I must calculate a jacobian matrix and check out the determinant, or something like that. If it's 1 then the volume of the cube is the same as the volume of this prism. And if it's 2 then the volume of the prism is half or twice the volume of the cube, etc. I don't know which jacobian to find though.
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  2. #2
    Member alunw's Avatar
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    Write the transformation as a matrix. That's very easy since you have been told what it does to the basis vectors. It is
    1 -2 0
    1 1 3
    0 1 -2
    (or the transpose if you want to write vectors using rows, so that you calculate the action by doing v*M rather than M*v)
    The determinant of the matrix is the volume of your transformed cube.
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by arbolis View Post
    T(C)=3^2.
    Can you explain how?
    T(C)=3^2
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  4. #4
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    You can do it directly. The three edges of C from (0,0,0) are given by the vectors <1, 0, 0>, <0, 1, 0>, and <0, 0, 1>. Applying T to each of those, we get, as we are told, <1, 1, 0>, <-2, 1, 1>, and <0, 3, -2>. Now use the fact that the volume of such a "parallelopiped" is given by the absolute value of the determinant
    \left|\begin{array}{ccc}1 & 1 & 0 \\ -2 & 1 & 1 \\ 0 & 3 & -2\end{array}\right| = \left|\begin{array}{cc}1 & 1 \\ 3 & -2\end{array}\right|- \left|\begin{array}{cc}-2 & 0 \\ 0 & -2\end{array}\right|= (-2-3)- 4= -9
    so the absolute value is 9= 3^2. The fact that a number happens to be a perfect square doesn't mean it must be an area!
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