A linear mapping $L_A :\mathbb{R}^3\rightarrow\mathbb{R}^3$ is given by $$A \cv{-1\\0\\1}=\cv{-4\\2\\4}, \quad A \cv{1\\1\\0}=\cv{1\\-1\\-1},\quad A \cv{0\\1\\2}=\cv{-5\\4\\4}\tiny.$$ We put this information in three columns with on top the column vector and at the bottom its image. For the sake of clarity, we place a horizontal line separating the top and bottom part: $$\left(\begin{array}{rrl} -1 & 1 & \phantom{-}0\\ 0 & 1 & \phantom{-}1\\ 1 & 0 & \phantom{-}2 \phantom{-}\\ \hline -4 & 1 & -5\\ 2& -1& \phantom{-}4\\ 4 & -1 & \phantom{-}4\end{array}\right)$$ We then transpose this matrix: $$\left(\,\begin{array}{rrr} -1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 2 \end{array}\,\left|\,\begin{array}{rrr} -4 & 2 & 4\\ 1 & -1 & -1 \\ -5 & 4 & 4 \end{array}\,\right.\right)$$ We reduce this matrix to it reduced row echelon form and get $$\left(\,\begin{array}{rrr} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\,\left|\,\begin{array}{rrr} 2 & 1 & -3\\ -1 & -2 & 2 \\ -2 & 3 & 1 \end{array}\,\right.\right)$$ We transpose this matrix and get $$\left(\begin{array}{rrl} 1 & 0 & \phantom{-}0\\ 0 & 1 & \phantom{-}0\\ 0 & 0 & \phantom{-}1\phantom{-}\\ \hline 2 & -1 & -2\\ 1 & -2 & \phantom{-}3\\ -3 & \phantom{-}2 & \phantom{-}1\end{array}\right)$$ By this construction appear in the columns on top the three unit vectors and at the bottom their image under $L_A$. So, the matrix $A$ is $$A=\left(\,\begin{array}{rrr} 2 & -1 & -2\\ 1 & -2 & 3\\ -3 & 2 & 1 \end{array}\,\right)$$ We can easily check whether there has been no calculation error by using this matrix to compute the images of $\cv{-1\\0\\1}, \cv{1\\1\\0}$ and $\cv{0\\1\\2}$ and to verify the results of this.

Of course you avoid the matrix transpositions by working with row vectors instead of column vectors in $\mathbb{R}^3$.