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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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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3,5,2
Could you tell me how you got that answer?
you have already mapped the first two

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Other answers:

the third one come from linear interpolation
this is just nomenclaturic bs, right?
It is part of a chapter called: Subspaces and linear transformations, in a book bought to calculus I
OK, 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.
so 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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