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Determinant of linear transformations

  1. Apr 26, 2009 #1
    I thought this problem was pretty straightforward, but I can't seem to match the answers in the back of the book.


    The problem is: Find the determinant of the following linear transformation.

    T(v) = <1, 2, 3> x v (where the x means cross product)

    from the plane V given by x + 2y + 3z = 0

    So I find the basis vectors

    <-2, 1, 0> and <-3, 0, 1>


    And I perform the transformation by

    T( <-2, 1, 0>) = <-3, 6, 5>
    T(<-3, 0, 1>) = < 2, 8, 6>


    And so I get the 2 column vectors to be

    | -3 2 |
    | 6 8 |
    | 5 6 |

    I know this is so off, but what am I doing wrong exactly?

    I can get the others but this one is giving me fits and I know it has to be so simple......
     
  2. jcsd
  3. Apr 26, 2009 #2
    Let u = <-2, 1, 0> and w = <-3, 0, 1>

    I haven't checked your calculations, so I'm assuming they're correct.

    The matrix representation you have for T is incorrect. First, you need to order your basis. Your basis will either be {u,w} or {w,u}. Let's say you choose the first ordered basis.

    What you did was say that T(u) = < -3, 6, 5> and so <-3, 6, 5> will be a column vector in the matrix. This is incorrect. You have to express <-3, 6, 5> as a linear combination of the basis vectors. So, T(u) = <-3, 6, 5> = a*u + b*w, where a and b are some scalars. Then <a,b> will be the first column vector for your matrix representation. Do the same thing with T(w) for some scalars c and d and so <c,d> will be your second column vector for the matrix, then just calculate the determinant.
     
  4. Apr 26, 2009 #3
    I see what you're saying. I had a brain fart. Thanks again.
     
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