Can You Add a Multiple of a Matrix Row to itself? I apologize in advance, as it seems this might be one of those questions that is so mind-boggling obvious to those who know the answer that most don't even think to treat upon it. I've searched up and down for information on matrice elementary row operations, but none of them have thought to explicitly treat upon the addition of a multiple of a matrix row to itself.
All of them say that it is possible to add a multiple of a matrix row to another row, but does adding a multiple of a matrix row to itself present a special case? Is it legal? It's a relatively straightforward question, I suppose, but I haven't been able to find confirmation one way or another on the topic. :(
Thank you very much for your time.
 A: Just left-multiply by 
$$\mathrm I + \alpha \, \mathrm e_i \mathrm e_i^{\top}$$
to multiply the $i$-th row by $1+\alpha$.
A: Each step of the Gauss elimination can be modeled by mulitplying the matrix with a so called elementary matrix, see here.

All of them say that it is possible to add a multiple of a matrix row
  to another row, but does adding a multiple of a matrix row to itself
  present a special case? Is it legal?

Sure it is legal, adding the $i$-th row of $A$ to row $i$ using the above definition would be the operation
$$
L_{i,i}(1) A
$$
with
$$
L_{i,i}(1) = \begin{bmatrix} 1 & & & & & & & \\ & \ddots & & & & & & \\ & & 2 & & & & & \\ & & & \ddots & & & & \\ & &  & & 1 & & \\ & & & & & & \ddots & \\ & & & & & & & 1\end{bmatrix}
$$
Note that this elementary matrix is the same as $D_i(2)$, scalar multiplying a row by $2$.
A: Yes, in the context of Gaussian elimination (which is where one allows adding multiples of one row to another row) there is an exception for addition to the same row; instead it is allowed to multiply a row by a nonzero constant. There is a good reason for the exception: if one were to allow adding a multiple of the row to itself, one could take the multiple to be $-1$ times the row, and the result would simply replace the rows by zero. That is not allowed, as it is a patently non-reversible operation; doing so will in general enlarge the solution set of the system under consideration. One could allow adding a multiple by any factor $\alpha\neq-1$, but that is already covered under the rule allowing multiplication by a nonzero constant (it has the same effect as multiplication by $1+\alpha$). Excluding $0$ as a multiplication factor is easier to state and to understand then excluding $-1$ as factor when adding a multiple, but only if the destination is the row itself.
