Consider the matrix of the transformation I need help with solving this hw problem for my matrix algebra class. 
Consider the transformation $T: \Bbb R^3 \to \Bbb R^2$ defined by:
$$
T(x) = T \pmatrix{x_1\\x_2\\ x_3} =
(2x_1 + x_3) \pmatrix{1\\2} + (x_2 - 3x_3)\pmatrix{-1\\1}
$$
1a) determine the matrix of the above transformation
1b) determine the reduced row echelon form of the matrix found in 1a
1c) based on your answer to part 1b, is the transformation T onto?
1d) based on your answer to part 1b, is the transformation T one-to-one?
1e) based on your answer to part 1b, determine the set of vectors x in R^3 for which T(x) = 0. Write your answer in parametric vector form
Would sincerely appreciate any help.
 A: Hint: For part a), it helps to rewrite
$$
T(x) = \pmatrix{2x_1 - x_2 + 4x_3\\ 4x_1 + x_2 - x_3}
$$
Now, for which matrix $A$ do we have $T(x) = Ax$?
If you go through the rest of the problem correctly, you should find that the transformation is onto, but not one-to-one.  Solving part e) is the same thing as finding the nullspace of the matrix.
A: For part (a), notice that 
$$T(x) = (2x_1 + x_3)\begin{bmatrix} 1\\2\end{bmatrix} + (x_2 -3x_3)\begin{bmatrix} -1\\1\end{bmatrix} = \begin{bmatrix} 2x_1 + x_3\\4x_1 +2x_3\end{bmatrix} +\begin{bmatrix} -x_2 +3x_3\\x_2 -3x_3\end{bmatrix}  = \begin{bmatrix} 2x_1 -x_2 + 4x_3\\4x_1 + x_2 -x_3\end{bmatrix} = \begin{bmatrix} 2 & -1 & 4\\4 & 1 & -1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}$$
Which implies that the matrix defining $T(x)$ is $$\begin{bmatrix} 2 & -1 & 4\\4 & 1 & -1 \end{bmatrix}$$
For 1e) I'm getting the reduced-form matrix as :
$$\begin{bmatrix}
2 & -1 & 4\\
4 & 1 &-1
\end{bmatrix}\rightarrow 
\begin{bmatrix}
1 & 0 & \frac{1}{2}\\
0 & 1 &-3
\end{bmatrix}
$$
Which gives the system of equations
$$
x_1 + \frac{1}{2}x_3 = 0
$$
$$
x_2 -3x_3 = 0
$$
Which implies that 
$$
x_1 = -\frac{1}{2}x_3
$$
$$
x_2 =3x_3
$$
Lastly, if we define $x_3 = t$, to solution in parametric form is then
$$
x_1 = -\frac{1}{2}t
$$
$$
x_2 =3t
$$
$$
x_3 =t
$$
