Check if there is a real linear mapping with properties $L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$

Question

I had a similar question before but this time I want make sure if I did it correctly myself on this example:

Check if there is a real linear mapping with properties $L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$

$L\begin{pmatrix} 1\\ 2 \end{pmatrix}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, \text{ }\text{ }L\begin{pmatrix} 2\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}, \text{ }\text{ }L\begin{pmatrix} -1\\ 4 \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$

If there exists such linear mapping, write its matrix form.

Let the matrix of the linear mapping $L$ be $M= \begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix}$

So for the first thing we have $\begin{pmatrix} a & b & c\\ d & e & f\\ g & h & i \end{pmatrix} \begin{pmatrix} 1\\ 2 \end{pmatrix}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$

and we already see that there is no linear mapping because the matrices have invalid size so we cannot even multiply them.

Is it correct like that?

Answer 1

$\begin{bmatrix} -1\\4\end{bmatrix} = 3\begin{bmatrix} 1\\2\end{bmatrix}-2\begin{bmatrix} 2\\1\end{bmatrix}$

If L were a linear map then $L\left(3\begin{bmatrix} 1\\2\end{bmatrix}-2\begin{bmatrix} 2\\1\end{bmatrix}\right) = 3L\begin{bmatrix} 1\\2\end{bmatrix} - 2L\begin{bmatrix} 2\\1\end{bmatrix}$

But it doesn't, so it is not.