1. ## Linear Transformation Problems

I have been stuck on a problem for awhile now and am in need of some help

Let $T:R^2 \rightarrow R^3$ be defined by $T(a_{1},a_{2}) = (a_{1}-a_{2},a_{1},2a_{1}+a_{2})$. Let $\beta$ be th standard ordered basis for $R^2$ and $\gamma$ = {(1,1,0),(0,1,1),(2,2,3)}. Compute $[T]^{\gamma}_{\beta}$.

Based on the answer in the back of the book it's supposed to be:

$[T]^{\gamma}_{\beta} = \begin{pmatrix} -1/3 & -1 \\ 0 & 1 \\ 2/3 & 0 \end{pmatrix}$

I have absolutely no idea how they got that solution, since when I input the values of the standard ordered basis I get $\begin{pmatrix} 1 & -1 \\ 1 & 0 \\ 2 & 1 \end{pmatrix}$.

2. Now, $\beta = \{ e_{1},e_{2} \}$, so $T( \beta ) = T(e_{1},e_{2}) = T \begin{pmatrix} 1 \\ 0 \end{pmatrix} , T \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Now, $T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = x \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + z \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}$, our job is to find x , y, and z.

Well, guess what? Plug in $- \frac {1}{3}, 0, \frac {2}{3}$ in for x, y, z, and see if they agree.