<s>[INST] Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6 \\ -14 \\ \end{bmatrix}$$ $$  T\Bigg (\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} -6\\ -40 \\ -2 \\ \end{bmatrix}
$$
Then the standard matrix for T is...
I'm not exactly sure how to approach this problem.
Could anyone explain how to solve this problem?
 [/INST] The standard matrix has columns that are the images of the vectors of the standard basis
$$
T \Bigg (\begin{bmatrix}1\\0\\0\end{bmatrix} \Bigg),  
\qquad
T \Bigg (\begin{bmatrix} 0\\1\\0 \end{bmatrix} \Bigg),
\qquad
T\Bigg (\begin{bmatrix}0\\0\\1 \end{bmatrix}\Bigg).
\tag{1}
$$
So one approach would be to solve a system of linear equations to write the vectors of the standard basis in terms of your vectors
$$
\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix},
\qquad
\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix},
\qquad
\begin{bmatrix}-4 \\ -5 \\ 5 \\ \end{bmatrix},
$$
and then obtain (1). 
Alternatively, note that if $A$ is the standard matrix you are looking for, then
$$
A \cdot 
\begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix}
=
\begin{bmatrix} 5 & -4 & -6\\ 3 & 6 & -40 \\ 14 & -14 & -2 \\ \end{bmatrix},
$$
and multiply on the right by the inverse of
$$
\begin{bmatrix}-2 & 3 & -4\\ 3 &-2&-5 \\ -4&3&5 \\ \end{bmatrix}.
$$
Spoiler And the matrix $A$ is...

$$\begin{bmatrix}-1& 5& 3\\ 5& 3&-1\\-2& -2 & -4\end{bmatrix}$$

Many Thanks to @MartinSleziak for correcting two misprints in comments below.
</s>