Let matrix
$$A = \begin{pmatrix} 1 & 2 & 0 \\ -3 & 1 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
and
$$B = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2 \end{pmatrix} $$
Find an elementary matrix $E$ such that $EA= B$.
I try with many
$$E = \begin{pmatrix} -1 & -2 & 0 \\ 1 & 2 & 0 \\ 3 & -1 & 1 \end{pmatrix} $$
but not correct
An elementary matrix is one which differs from the identity matrix by one elementary row operation.
Note that $B$ is the matrix $A$ with three times the first row added to the second. So if we take the matrix
$$E=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$
and now consider
$$EA=\begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 2 & 0 \\ -3 & 1 & 1 \\ 0 & 4 & 2\end{pmatrix}=\begin{pmatrix}1 & 2 & 0 \\ 0 & 7 & 1 \\ 0 & 4 & 2\end{pmatrix}=B$$
$E$ was the identity matrix with three times the first row added to the second. This is easy to figure out because the matrix $B$ is linear combinations of the rows of $A$ and linear combinations of the columns of $E$.