Logic of Elementary Row Operations to Create Equivalent Systems Can anyone explain why the 3rd operation applied on a system creates an equivalent system with the same solution. 
Elementary Row Operations.
1.
Interchange two rows.
2.
Multiply a row with a nonzero number.
3.
Add a row to another one multiplied by a number. 
 A: Think about it as if you are dealing with a system of equations. For example, $$x+2y=0\\3x+4y=4$$When you multiply equation $1$ by a certain number, let's say $-3$, the equation becomes $-3x-6y=0$, which does not change any value of the variables. 
Then add this new equation to the second equation, now they become $-2y=4$, and you can now solve for $x$ and $y$.
So it is basically the same process as you solve for a system of equations with multiple variables, this is just changed into the matrix form.
A: a system of linear equations is equivalent to an matrix equation
of type $ Ax = a$, then the three operations indicated are the
transformation of the matrix equation  $ Ax = b$ in the matrix
equation  $PAx = Pa$ where $P$ is an invertible matrix and so the
two matrix equation have the same solutions.
trasformation in step (1) $R_i$ to $R_j$: that is $P = (a_ {kl})$ where
$a_{ij}=a_{ji}=a_ {kk}=1$ for all $k$,  $i\not=k\not=j$ and  $a_
{kl}=0$ for all others $k, l$.
in step (2) $R_i$ to $\lambda R_i$ with $\lambda\not=0$: that is $P =
(a_ {kl})$ where $a_{ii}=\lambda$,  $a_{jj}=1, j\not=i$  and $a_
{kl}=0$ for all others $k, l$.
in step (3) $R_i $ to $R_i+\lambda R_j$: that is $P = (a_ {kl})$
where $a_{ij}=\lambda$,  $a_{kk}=1$, and $a_ {kl}=0$ for all
others $k, l$.
