Let T : $\mathbb{R^3}$$\to$$\mathbb{R^3}$ be the linear transformation $$T(x_1,x_2,x_3) = (2x_1-x_2,2x_2+3x_3,3x_1+4x_3)^T$$ for all $(x_1,x_2,x_3) \in \mathbb{R^3}$
I need to calculate $[T]_\varepsilon$ of the operator $T$ corresponding to the standard basis $\varepsilon$ of $\mathbb{R^3}$, where $\varepsilon={\{e_1,e_2,e_3\}}$, $e_1=(1,0,0)^T , e_2=(0,1,0)^T , e_3=(0,0,1)^T$.
I also need to show whether $T$ is one-to-one or not, and explain why.
What I did so far (not sure if I am correct):
$[T(e)]_\varepsilon = A[e]_\varepsilon$ $= Ae_1 + Ae_2 + Ae_3$ $$=\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}1\\0\\0\\\end{array}}\right]+\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}0\\1\\0\\\end{array}}\right]+\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}0\\0\\1\\\end{array}}\right]= A$$
I have just gone in a circle and come back out with A.