$L:\mathbb{R}^3\rightarrow \mathbb{R}^2$ with bases
$\mathcal{S}=\left\{\left(-1,1,0\right),\left(0,1,1\right),\left(1,0,0\right)\right\} \: \text{for} \:\mathbb{R}^3 \:\text{and} \\ \mathcal{T}=\left\{\left(1,2\right),\left(1,-1\right)\right\}\: \text{for} \:\mathbb{R}^2.\:$
The matrix of $L$ with respect to $\mathcal{S}$ and $\mathcal{T}$ is: $A=\begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 0 \end{bmatrix}$
I used the formula $A\bf{x}$ $= \bf{b}$ to compute $\left[L\left(v_1\right)\right]_\mathcal{T}$ and ended up with $\begin{bmatrix}1 \\ 2\end{bmatrix}^\mathcal{T}$, but this answer is wrong according to the study guide I'm working from. The answer it has is $\begin{bmatrix}1\\ -1\end{bmatrix}^\mathcal{T}$. What am I doing wrong?
my work:$\begin{pmatrix}1 & 2 & 1 \\-1 & 1 & 0\end{pmatrix}\cdot \begin{pmatrix}-1 \\1 \\0\end{pmatrix}\:=\:\begin{pmatrix}1 \\2\end{pmatrix}$
The way you have things defined, it looks like you can read off the column vectors you seek from that matrix with respect to the linear transformation. The matrix of $L$ with respect to $\mathcal{S}$ and $\mathcal{T}$ is one that is precisely $$\begin{bmatrix}[L\mathbf{v_1}]_{\mathcal{T}} & [L\mathbf{v_2}]_{\mathcal{T}} & [L\mathbf{v_3}]_{\mathcal{T}}\end{bmatrix}$$ where $(\mathbf{v_1},\mathbf{v_2},\mathbf{v_3})$ is a basis for $\mathbb{R}^3$, and $[L\mathbf{v_j}]_{\mathcal{T}}$ is a column vector of the coordinates of $L\mathbf{v_j}$ with respect to the basis $\mathcal{T}$ of $\mathbb{R}^2$. The work is already done for you.