We have

\[\begin{bmatrix}1 & \lambda \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}+\begin{bmatrix}0 & \lambda \\0 & 0 \end{bmatrix}=I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}\]

It is the position \(\displaystyle_{\left\lbrace{12}\right\rbrace}\) of the blue marked 1 which determines selected row resp. column of A.

We obtain

\[Ae_{12}(\lambda)=A(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})=A+\lambda\begin{bmatrix}0 & a \\0 & c \end{bmatrix}\]

\[e_{12}(\lambda)A=(I+\lambda\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix})A=A+\lambda\begin{bmatrix}c & d \\0 & 0 \end{bmatrix}\]