Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
So I have forgotten how to do this. Do I:
- reduce $A$ to reduced row-echelon form (why do I do this? Is it because it's easy to solve once you have pivot columns and free variables)?
- rewrite the system of equations
Is this right:
\begin{align}
A &= \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & -5 & -2 & 7 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{-29}{5} \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix}
\end{align}
so: $v_4 = t, v_3 = s, v_2 = \frac{-2}{5}s - \frac{7}{5}t, v_1 = \frac{-1}{5}s + \frac{29}{5}t$
$$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{5}s + \frac{29}{5}t \\ \frac{-2}{5}s - \frac{7}{5} t \\ s + 0t \\ 0 + t \end{bmatrix} = s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix} $$
Is this the set of all $v$?
EDIT
I messed up, the first $\frac{7}{5}$ should be a $\frac{-7}{5}$

Hint: Augment the matrix $A$ with the column $\begin{pmatrix} 1\\2\end{pmatrix}$ and row-reduce the resulting augmented matrix.   
So, row reduce $\left (\begin{array}{rrrr|r}2&-1&0&1&1\\1&2&1&-3 &2\end{array}\right ) $.
