Argthtjtr

Many linear algebra courses build upon the concept of elementary row operations, and further develop on their properties and relevance.

My question is, why isn't addition by a real considered an elementary row operation and how can this be argued at the level of the beginning of the course? I know that this operation doesn't share some properties of the elementary row operations. However, linear algebra courses present elementary row operations as operations that don't change the solution set. Well, addition with a real doesn't change the solution set.

Take for example the augmented matrix
$$(begin{array}{cc|c} 1 & -1 & 0 end{array}).$$
This has a solution of $(x, y) = t(1, 1)$.

Add the real number $1$ to the row:
$$(begin{array}{cc|c} 2 & 0 & 1 end{array}).$$
This has a solution of $(x, y) = (1/2, 0) + t(0, 1)$, which is a completely different line of solutions.

So, adding a real does indeed change the solution set.

More generally, there are other operations that do preserve the solution set, but only three are called "elementary". The reason why we restrict ourselves to just three operations is because these three operations are sufficient to transform between any two matrices of the same dimensions with the same solution set. Every row operation that preserves solution sets is a combination of elementary row operations!

Expanding on the first part of Theo Bendit's answer: if we represent the equation $$ax+by=c$$ by the matrix row $$(begin{array}{cc|c} a & b & c end{array})$$
and add a constant $k$ to the row, the result
$$(begin{array}{cc|c} a+k & b+k & c+k end{array})$$
is equivalent to the equation

$$(a+k)x+(b+k)y=c+k.$$

Comparing with the original equation, we see we've added $k(x+y)$ to one side and $k$ to the other—producing a completely different equation.

So adding a constant isn't a valid row operation.

The elementary row operations are supposed to correspond to left multiplication by an invertible matrix. Some examples on a $3 times 3$ matrix.

1. Row switching:

$qquad$ Switching rows $1$ and $2$ corresponds to left multiplication by
$ T = left(begin{array}{c}
0 & 1 & 0 \
1 & 0 & 0 \
0 & 0 & 1 \
end{array} right)$
and $ T^{-1} = T$.

2. Multiply all elements of a row by a constant:

$qquad$ Multiplying row 2 by $m ne 0$ corresponds to left multiplication by
$T = left(begin{array}{c}
1 & 0 & 0 \
0 & m & 0 \
0 & 0 & 1 \
end{array} right)$

$qquad$ and $T^{-1} = left(begin{array}{c}
1 & 0 & 0 \
0 & frac 1m & 0 \
0 & 0 & 1 \
end{array} right)$.

3. Add row $i$ multiplied by a scalar $m$ to row $j$.

$qquad$ Adding $m$ times row $1$ to row $2$ corresponds to left multiplication by
$T = left(begin{array}{c}
1 & 0 & 0 \
m & 1 & 0 \
0 & 0 & 1 \
end{array} right)$

$qquad$ and $T^{-1} = left(begin{array}{c}
1 & 0 & 0 \
-m & 1 & 0 \
0 & 0 & 1 \
end{array} right)$.

There is no such matrix that corresponds to adding a constant to a row.