Sorry, the title is not as descriptive as I wanted it to be.
The proposition I would like to prove:
Consider arbitrary $n$ by $n$ matrix $\bf A$ matrix and $n$ by $n$ elementary matrix $\bf E$ such that
$$\mathbf{E}_{i,j} = \begin{cases} \mathbf{I}_{i,j}, & i ≠ r\\[2ex] \mathbf{I}_{i,j} + k\mathbf{I}_{d,j}, & i = r \\ \end{cases}$$
Prove that
$$[\mathbf{EA}]_{i,j} = \begin{cases} \mathbf{A}_{i,j}, & i ≠ r\\[2ex] \mathbf{A}_{i,j} + k\mathbf{A}_{d,j}, & i = r \\ \end{cases}$$
Note that that $1≤r≤n$ and $1 ≤ d ≤ n$ and $k$ is a scalar.
My attempt:
Take arbitarary $[\mathbf {EA}]_{i,j}$, $i ≠ r$
We have
$$\begin{align}[\mathbf{EA}]_{i,j} & = \sum_{x = 1}^{n}\mathbf{E}_{i,x}\mathbf{A}_{x,j} \\ & = \sum_{x = 1}^{n}\mathbf{I}_{i,x}\mathbf{A}_{x,j} = \mathbf{A}_{i,j} \end{align}$$
Now take arbitarary $[\mathbf {EA}]_{i,j}$, where $i = r$
$$\begin{align}[\mathbf{EA}]_{i,j} & = \sum_{x = 1}^{n}\mathbf{E}_{i,x}\mathbf{A}_{x,j} \\ & = \sum_{x = 1}^{n}[\mathbf{I}_{i,x} + k\mathbf{I}_{d,x}]\mathbf{A}_{x,j} \\ & = \sum_{x = 1}^{n}\bigl(\mathbf{I}_{i,x}\mathbf{A}_{x,j} + k\mathbf{I}_{d,x}\mathbf{A}_{x,j}\bigr) \\ & = \sum_{x = 1}^{n}\mathbf{I}_{i,x}\mathbf{A}_{x,j} + k\sum_{x = 1}^{n} \mathbf{I}_{d,x}\mathbf{A}_{x,j} \\ & = \mathbf{A}_{i,j} + k\mathbf{A}_{d,j} \end{align}$$
As desired. $\square$
Is it correct?