linear transformation $\Bbb R^3 \to \Bbb R^2$ For linear transform $T ( 1,1,1 ) = (1 , 2),\,\, T ( 1, 1, 0) = (2 , 3)$ and $\,\,T ( 1,0,0 ) = ( 3 , 4)$,from $\mathbb{R}^3$ to $\mathbb{R}^2$, the function used was : Choose one : 
a. $T ( x, y, z) = (x - y - 3z , 4x - y - 2z )$ 
b. $T ( x, y, z) = (3x + y + z, 4x + y + z)$ 
c. $T ( x, y, z) = (3x - y - z, 4x - y -z )$
d. $T ( x, y, z) = (x - 3y - z, 4x - y -z )$
e. $T ( x, y, z) = (3x + y - z, 4x - y + z)$
 A: We need to find the $T$ from our choices (a) through (e) such that
$$
T\begin{bmatrix}
1\\1\\1
\end{bmatrix} = \begin{bmatrix}
1\\2
\end{bmatrix}, \ T\begin{bmatrix}
1\\1\\0
\end{bmatrix} = \begin{bmatrix}
2\\3
\end{bmatrix}, \ T\begin{bmatrix}
1\\0\\0
\end{bmatrix}= \begin{bmatrix}
3\\4
\end{bmatrix}.
$$
Since we know three inputs and what their outputs should be, we can just directly check which $T$ is the correct one. So we will take our inputs $(1,1,1)$, $(1,1,0)$, and $(1,0,0)$ and plug them into each of the possibilities (a) through (e). If the output isn't correct (the correct ones we listed above), then we can mark that $T$ off our list.
For instance, we know that (a) can't be the correct one since
$$
T\begin{bmatrix}
1\\1\\1
\end{bmatrix} = \begin{bmatrix}
1-1-3\cdot 1 \\
4\cdot 1 - 1 - 2\cdot 1
\end{bmatrix} = \begin{bmatrix}
-3 \\
1
\end{bmatrix}
$$
but we need to have $T(1,1,1)=(1,2)$. Now you can plug $(1,1,1)$ into the remaining through possible $T$'s. If that doesn't tell you which is the correct one, you repeat the same process with $(1,1,0)$. If that doesn't eliminate all possibilities, you finally work through $(1,0,0)$.
A: The answer is (c). One can rewrite the given equations in matrix form as follows:
$$ T \cdot\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix}.$$
Hence, we have 
$$ T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \cdot\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & -1 & -1 \\ 4 & -1 & -1 \end{bmatrix}.$$
