## 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 $T(1,3,-4)$

1. IrishBoy123

3,5,2

2. anonymous

Could you tell me how you got that answer?

3. IrishBoy123

you have already mapped the first two

4. IrishBoy123

the third one come from linear interpolation

5. IrishBoy123

this is just nomenclaturic bs, right?

6. anonymous

It is part of a chapter called: Subspaces and linear transformations, in a book bought to calculus I

7. IrishBoy123

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.

8. IrishBoy123

so you can apply these rules to find the transformation of the third vector $$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!$$ $$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \!$$