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?

You can put into a matrix given vectors and their images. If you then do elementary row operations, this property is not changed. (After each step you have in each row a vector and its image. This is because of linearity.)
If you manage to obtain the identity matrix on the left, then you know the images of the vectors from the standard basis, which is sufficient to obtain the matrix of your linear transformation. (Since you're using column vectors, the result is the transpose of the matrix on the right. It is worth mentioning that some authors prefer row vectors - in such case the matrix wouldn't be transposed. See also the Wikipedia article on Row and column vectors.)
$
\left(\begin{array}{ccc|ccc}
-2 &  3 & -4 &  5 & 3 & 14\\
 3 & -2 &  3 & -4 & 6 &-14\\
-4 & -5 &  5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  1 & -1 &  1 & 9 & 0\\
 3 & -2 &  3 & -4 & 6 &-14\\
-4 & -5 &  5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  1 & -1 &  1 & 9 & 0\\
 0 & -5 &  6 & -7 &-21 &-14\\
-4 & -5 &  5 & -6 &-40&-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  1 & -1 &  1 &  9 & 0\\
 0 & -5 &  6 & -7 &-21 &-14\\
 0 & -1 &  1 & -2 & -4 &-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  0 &  0 & -1 &  5 &-2\\
 0 & -5 &  6 & -7 &-21 &-14\\
 0 & -1 &  1 & -2 & -4 &-2
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  0 &  0 & -1 &  5 &-2\\
 0 &  1 & -1 &  2 &  4 & 2\\
 0 & -5 &  6 & -7 &-21 &-14
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  0 &  0 & -1 &  5 &-2\\
 0 &  1 & -1 &  2 &  4 & 2\\
 0 &  0 &  1 &  3 & -1 &-4
\end{array}\right)\sim
\left(\begin{array}{ccc|ccc}
 1 &  0 &  0 & -1 &  5 &-2\\
 0 &  1 &  0 &  5 &  3 &-2\\
 0 &  0 &  1 &  3 & -1 &-4
\end{array}\right)
$
