Finding an unknown linear transformation given that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$ 
Let $T\colon\mathbb{R}^2\to\mathbb{R}^3$ be the linear transformation such that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$. 
  

Does such a linear transformation exist?
So far I've worked out that it cannot exist, as the first entry for $T(1,1)$ is 1, whereas the first entry for $T(2,3)$ is 1, which can only be obtained through 3-2, or half of 2.
Is there any way of presenting this formally?
 A: Hint: $\{(1,1),(2,3)\}$ is a basis of $\mathbb R^2$.
A: I don't think that your reason works. If a linear transformation exists, it should be a 2 by 3 matrix. Let this matrix have the entries $a$ and $b$ on the first row, $c$ and $d$ on the second and $e$ and $f$ on the third.
Performing matrix multiplication on <1,1> and <2,3> to get <1,0,2> and <1,-1,4> gives the following systems of equations to solve: $a+b=1$ with $2a+3b=1$, and $c+d=0$ with $2c+3d=-1$  lastly $e+f=2$ with $2e+3f=4$ These systems produce unique values for the matrix' entries. Sorry for my poor formatting
A: Spoiler Warning: Full Solution
We can write this as a matrix vector equation $${\bf T}\left[\begin{array}{rr}1&2\\1&3\end{array}\right] = \left[\begin{array}{rrr}
1&1\\
0&-1\\
2&4
\end{array}\right]$$
Since Arpit mentioned in the hint, $\{(1,1),(2,3)\}$ spans a basis for $\mathbb{R}^2$, why the matrix must be invertible. Now if we multiply each side to the right with $\left[\begin{array}{rr}1&2\\1&3\end{array}\right]^{-1}$:
$${\bf T} = \left[\begin{array}{rrr}
1&1\\
0&-1\\
2&4
\end{array}\right] \left[\begin{array}{rr}1&2\\1&3\end{array}\right]^{-1} = \left[\begin{array}{rr}
2&-1\\
1&-1\\
2&0
\end{array}\right]$$
Now what remains is to confirm that the first equation is satisfied.
A: Hint: A linear transformation $T: V \to W$ is uniquely determined by how it acts on the elements of a basis for $V$.
A: There is a very simple method to solve the problem described in "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely."
Consider the formula
$$
\vec{T}(\vec{p}) = (-1)
\frac{
    \det
        \begin{pmatrix}
            0   & \vec{x} & \vec{y} \\
            p_1 & a_1     & b_1     \\
            p_2 & a_2     & b_2     \\
        \end{pmatrix}
}{
    \det
        \begin{pmatrix}
            a_1 & b_1 \\
            a_2 & b_2 \\
        \end{pmatrix}
},
$$
where $\vec{T}$ is linear transformation acting on arbitrary point $\vec{p}$. $\vec{T}$ has the property
$$
\vec{T}(\vec{a}) = \vec{x};\quad
\vec{T}(\vec{b}) = \vec{y}.
$$
Indices designate components of the corresponding vector.
Let's consider your case.
We need such $\vec{T}$ that
$$
\vec{T}: \begin{pmatrix}1\\ 1\end{pmatrix} \mapsto 
         \begin{pmatrix}1\\ 0\\ 2\end{pmatrix};~
\vec{T}: \begin{pmatrix}2\\ 3\end{pmatrix} \mapsto 
         \begin{pmatrix}1\\-1\\ 4\end{pmatrix}.
$$
Now I plug them into the general expression 
$$
\vec{T}(\vec{p}) =
(-1)
\frac{
    \det
    \begin{pmatrix}
        0 & (1,0,2)^T & (1,-1,4)^T\\
        \begin{matrix}
            p_{1} \\
            p_{2}
        \end{matrix} &
%
        \begin{matrix}1\\ 1\end{matrix} &
%
        \begin{matrix}2\\ 3\end{matrix}
    \end{pmatrix}
}{
    \det
    \begin{pmatrix}
        \begin{matrix}1\\ 1\end{matrix} &
%
        \begin{matrix}2\\ 3\end{matrix}
    \end{pmatrix}
}.
$$
Doing determinants I get
$$
\vec{T}(\vec{p})
=   \left[
   3\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} -
    \begin{pmatrix}1\\-1\\ 4\end{pmatrix}
\right] p_1 -
  \left[
   2\begin{pmatrix}1\\ 0\\ 2\end{pmatrix} -
    \begin{pmatrix}1\\-1\\ 4\end{pmatrix}
\right] p_2
$$
or simplified
$$
\vec{T}(\vec{p}) =
    \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} p_1 + 
    \begin{pmatrix} -1 \\ -1\\ 0 \end{pmatrix} p_2. 
$$
Of course, you can write that in a vector form
$$
\vec{T}(\vec{p}) =
    \begin{pmatrix} 
        2 &-1 \\
        1 &-1 \\
        2 & 0 
    \end{pmatrix} 
    \begin{pmatrix} p_1 \\ p_2 \end{pmatrix}.
$$
Now you can easily check
$$
    \begin{pmatrix} 
        2 &-1 \\
        1 &-1 \\
        2 & 0 
    \end{pmatrix} 
    \begin{pmatrix}1\\ 1\end{pmatrix} =
    \begin{pmatrix}1\\ 0\\ 2\end{pmatrix};~
    \begin{pmatrix} 
        2 &-1 \\
        1 &-1 \\
        2 & 0 
    \end{pmatrix} 
    \begin{pmatrix}2\\ 3\end{pmatrix} =
    \begin{pmatrix}1\\ -1\\ 4\end{pmatrix}.
$$
For more details on the methods used, you can always refer to "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely". The latter contains many problems similar to this one as explained by the authors of the method presented.
