I have tried to solve the following exercise. Is it right?
Consider the linear transformation L: ℝ⁴→ ℝ³. Knowing that:
$$ L \begin{pmatrix}2\\0\\0\\0\end{pmatrix} = \begin{pmatrix}2\\2\\2\end{pmatrix} \space L \begin{pmatrix}1\\1\\0\\0\end{pmatrix} = \begin{pmatrix}1\\1\\0\end{pmatrix} \space L \begin{pmatrix}1\\1\\1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\end{pmatrix} \space L \begin{pmatrix}1\\1\\1\\1\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix} \space $$ Determine the image of the elements in the standard basis of ℝ⁴ through L.
$$ \begin{pmatrix}1\\0\\0\\0\end{pmatrix} = \frac{1}{2} \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \\ L \begin{pmatrix}1\\0\\0\\0\end{pmatrix} = L \left( \frac{1}{2} \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \right) = \frac{1}{2} \begin{pmatrix}2\\2\\2\end{pmatrix} = \begin{pmatrix}1\\1\\1\end{pmatrix} $$
$$ \begin{pmatrix}0\\1\\0\\0\end{pmatrix} = -\frac{1}{2} \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \\ L \begin{pmatrix}0\\1\\0\\0\end{pmatrix} = L \left( -\frac{1}{2} \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \right) = -\frac{1}{2} \begin{pmatrix}2\\2\\2\end{pmatrix} + 1 \begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}0\\0\\-1\end{pmatrix} $$
$$ \begin{pmatrix}0\\0\\1\\0\end{pmatrix} = 0 \begin{pmatrix}2\\0\\0\\0\end{pmatrix} - 1 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \\ L \begin{pmatrix}0\\0\\1\\0\end{pmatrix} = L \left( 0 \begin{pmatrix}2\\0\\0\\0\end{pmatrix} - 1 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \right) = -1 \begin{pmatrix}1\\1\\0\end{pmatrix} + 1 \begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}0\\-1\\0\end{pmatrix} $$
$$ \begin{pmatrix}0\\0\\0\\1\end{pmatrix} = 0 \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} - 1 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \\ L \begin{pmatrix}0\\0\\0\\1\end{pmatrix} = L \left( 0 \begin{pmatrix}2\\0\\0\\0\end{pmatrix} + 0 \begin{pmatrix}1\\1\\0\\0\end{pmatrix} - 1 \begin{pmatrix}1\\1\\1\\0\end{pmatrix} + 1 \begin{pmatrix}1\\1\\1\\1\end{pmatrix} \right) = - 1 \begin{pmatrix}1\\0\\0\end{pmatrix} + 1 \begin{pmatrix}0\\0\\0\end{pmatrix} = \begin{pmatrix}-1\\0\\0\end{pmatrix} $$
Your answer is right, but your method looks a bit convoluted. In these cases, one usually works with matrices that represent the transformation. First, determine the change of basis matrix $M_{\mathcal{S}_4,\mathcal{B}}$, where $\mathcal{B} \subset \mathbb{R}^4$ is the given basis and $\mathcal{S}_4 \subset \mathbb{R}^4$ is the standard basis. Then the linear transformation matrix is $$ M_{\mathcal{S}_4,\mathcal{S}_3}(L) = M_{\mathcal{B},\mathcal{S}_3}(L) \cdot M_{\mathcal{S}_4,\mathcal{B}} $$ We know that $$M_{\mathcal{S}_4,\mathcal{B}}=M_{\mathcal{B},\mathcal{S}_4}^{-1} = \begin{pmatrix} 2 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}^{-1}= \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ and $$ M_{\mathcal{B},\mathcal{S}_3}(L) = \begin{pmatrix} 2 & 1 & 1 & 0 \\ 2 & 1 & 0 & 0 \\ 2 & 0 & 0 & 0 \end{pmatrix} $$ Where $\mathcal{S}_3$ is the standard basis in $\mathbb{R}^3$. Therefore, $$ M_{\mathcal{S}_4,\mathcal{S}_3}(L) = \begin{pmatrix} 2 & 1 & 1 & 0 \\ 2 & 1 & 0 & 0 \\ 2 & 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & -1 \\ 1 & 0 & -1 & 0 \\ 1 & -1 & 0 & 0 \end{pmatrix} $$ The columns of $M_{\mathcal{S}_4,\mathcal{S}_3}$ are the images of the vectors of $\mathcal{S}_4$.