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 \(\mathbb{R}^3\).