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anonymous
 one year ago
The linear transformation \[T:R^3>R^3\] depicts the vector\[u=(2,3,1)\] in \[T(u)=(4,5,2)\] and the vector \[v=(1,1,2)\] in \[T(v)=(3,1,2)\] Find \[T(1,3,4)\]
anonymous
 one year ago
The linear transformation \[T:R^3>R^3\] depicts the vector\[u=(2,3,1)\] in \[T(u)=(4,5,2)\] and the vector \[v=(1,1,2)\] in \[T(v)=(3,1,2)\] Find \[T(1,3,4)\]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you tell me how you got that answer?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0you have already mapped the first two

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0the third one come from linear interpolation

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0this is just nomenclaturic bs, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is part of a chapter called: Subspaces and linear transformations, in a book bought to calculus I

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0OK, gotcha first \[\left\begin{matrix}2 & 3 &1 \\ 1 & 1 & 2 \\ 1 & 3 & 4\end{matrix}\right = 0\] which makes you wonder then a bit of row reductions shows that \[2\left[\begin{matrix}2 \\ 3\\1\end{matrix}\right]  3\left[\begin{matrix}1 \\ 1\\2\end{matrix}\right] = \left[\begin{matrix}1 \\ 3\\4\end{matrix}\right]\] so they are linearly dependent vectors.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0so you can apply these rules to find the transformation of the third vector \(f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!\) \(f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \!\)
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