Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that 
T $\begin{pmatrix} 
        1 \\
        0 \\ 
        0 \\
        \end{pmatrix}$ = $\begin{pmatrix} 
        3 \\
        1 \\ 
        4 \\
        \end{pmatrix}$, $T$ $\begin{pmatrix} 
        0 \\
        1 \\ 
        0 \\
        \end{pmatrix}$ = $\begin{pmatrix} 
        -1 \\
        -1\\ 
        3\\
        \end{pmatrix}$ , $T$ $\begin{pmatrix} 
        0 \\
        0 \\ 
        1 \\
        \end{pmatrix}$ = $\begin{pmatrix} 
        4 \\
        -3 \\ 
        -1\\
        \end{pmatrix}$ then $T$ $\begin{pmatrix} 
        -5\\
        4 \\ 
        4 \\
        \end{pmatrix}$ = $\begin{pmatrix} 
        ? \\
        ? \\ 
        ? \\
        \end{pmatrix}$

I do not know where to start with this question, I tried doing RREF of the transformation numbers but I feel that is wrong. 
 A: Define $\vec e_i = (0, \dots, 1, 0, \dots)$ where the 1 is in the $i^\text{th}$ position. Then note that your question is
$$ T(-5\vec e_1 + 4 \vec e_2 + 4 \vec e_3) = -5\pmatrix{3 \\ 1 \\ 4} + 4\pmatrix{-1 \\ -1 \\ 3} + 4\pmatrix{4 \\ -3 \\ -1} = \pmatrix{-3 \\ -21 \\ -12}.$$
A: To expand on my comment above since @Gregory already gave the answer using one method:
Given information on how $T$ acts on the standard basis, knowing $T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}3\\1\\4\end{bmatrix}$ and so on, this tells you exactly how to represent $T$ as a matrix.  The columns of $T$ are quite simply the corresponding results of $T$ being applied to the standard basis vectors.
We get then $T=\begin{bmatrix}3&-1&4\\1&-1&-3\\4&3&-1\end{bmatrix}$ and so all that remains to your problem is completing the matrix arithmetic necessary.
A: Let $\{ e_1,e_2,e_3 \}$ be the canonical (standard) basis of $\mathbb{R}^3$. Note that for any $(x,y,z) \in \mathbb{R}^3$ we can write $(x,y,z)=ze_1+ye_2+ze_3$. Hence $T(x,y,z)=T(xe_1+ye_2+ze_3)$, and using lineality $$T(x,y,z)=T(xe_1)+T(ye_2)+T(ze_3)=xT(e_1)+yT(e_2)+zT(e_3)$$ and by hypotesis $$T(x,y,z)=x(3,1,4)+y(-1, -1, 3)+z(4,-3,-1)=(3x-y+4z, x-y-3z, 4x+3y-z).$$
Finally, $$T(-5,4,4)=(-15-4+16,-5-4-12, -20+12-4)=(-3, -21, -12).$$
