Given is $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with $f:(x,y,z) \rightarrow (x+2y+z, y+z, -x+3y+4z)$. Determine the transformation matrix in terms of the basis $B= \left\{\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix},\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}\right\}$
$f(\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}= 0 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}- 1 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
$f(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix})= \begin{pmatrix} 3\\ 2\\ 7 \end{pmatrix}= 8 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+5 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
$f(\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix})= \begin{pmatrix} 0\\ 1\\ 5 \end{pmatrix}= 4 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 1 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+ 4 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
I'm not sure how to read the transformation matrix now. Either I will read it correctly or I will accidentally read its transposition :s
$T=\begin{pmatrix} 0 & 8 & 4\\ 0 & 2 & 1\\ -1 & 5 & 4 \end{pmatrix}$
Did I do it all correctly? If it's alright, there are maybe faster ways doing this?
Perhaps a more straightforward way of computing the matrix is by using the relation \begin{equation} [T]_\mathcal{B} = P^{-1} T P, \end{equation} where $[T]_\mathcal{B}$ is the matrix that you computed, $$P = (b_1\quad b_2\quad b_3) = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}, $$ and $$T = \begin{pmatrix} f(e_1) & f(e_2) & f(e_3) \end{pmatrix} = \begin{pmatrix}1 & 2 & 1\\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{pmatrix},$$ where $e_1, e_2, e_3$ are the standard basis vectors. With this approach, all you have to do is invert $P$, which is easy, and then multiply the matrices. (check to see that it yields the same result!)
Heres a nice diagram that illustrates why the relation is true, from Linear Algebra Basis Trick.