Question about reducing a matrix to Row Echelon form

Question

First of all I want to apologize since this question is actually a bit too elementary. Given the following matrix $A$, I want to reduce it to its row echelon form: $$A=\begin{bmatrix}2&0&a&2\\-1&0&-1&3\\5&a+4&-4&-3\end{bmatrix}$$

Let's say I want to zero-out the element $a_{12}=-1$. For this, one would usually do: $$\text{Row}_{(2)}\longrightarrow \text{Row}_{(2)}+\dfrac12\text{Row}_{(1)}$$

But one could also do this (I think): $$\text{Row}_{(2)}\longrightarrow 2\times\text{Row}_{(2)}+\text{Row}_{(1)}$$

Again apologize for such a basic question, but I don't know if I'm allowed to multiply the same row I'm operating on (in this case, row 2) by a factor. I always thought that it had to be a linear combination of other rows EXCEPT the row I'm manipulating.

In this case, if I'm operating on row 2, can I multiply it by a factor and sum it to other rows etc? In other words, is the second operation above correct/allowed?

Does this also hold when I'm discussing a system of equations in terms of 1 or 2 coefficients, and I have to reduce the coefficients matrix to its row-echelon form?

Thank you for your time and patience.

Answer 1

Yes, you can do that if the factor that you are multiplying by is not $0$.

Answer 2

See Elementary row operations. According to the definition, $$\text{Row}_{(2)}\longrightarrow 2\times\text{Row}_{(2)}+\text{Row}_{(1)}$$ is the same as a row multiplication, followed by a row addition: $$\text{Row}_{(2)}\longrightarrow 2\times\text{Row}_{(2)}$$ $$\text{Row}_{(2)}\longrightarrow \text{Row}_{(2)}+\text{Row}_{(1)}.$$ Therefore, you're right.

So $$\text{Row}_{(2)}\longrightarrow 2 \times\text{Row}_{\color{blue}{(3)}}+\text{Row}_{\color{blue}{(1)}} $$ is not allowed.