TITLE: Are column operations legal in matrices also?
QUESTION [9 upvotes]: In linear algebra we have been talking a lot about the three elementary row operations. What I don't understand is why we can't multiply by any column by a constant? Since a matrix is really just a grouping of column vectors, shouldn't we be able to multiply a whole column by a constant but maintain the same set of solutions for the original and resulting matrices?

REPLY [5 votes]: I want to point out that you can solve a linear system with column operations. Column operations do change the solution... but the new solution is in terms of different variables! A column operation is a change of variables!
(This is unsurprising of course; row operations corresponds to multiplying by elementary matrices on the left, $\cdots E_2 E_1 A x = y$, ie. we're transforming the result of applying $A$, while column operations correspond to multiplying on the right, $A E_1 E_2\cdots x = y$, ie. we're transforming the variables before they get fed into $A$.)
If you have $\begin{align}2X+3Y&=1\\6X-4Y&=2\end{align}$

Multiplying the first column by $k$ corresponds to the substitution $X = kX'$.
$$\begin{align}2kX'+3Y&=1\\6kX'-4Y&=2\end{align}$$
Exchanging the columns corresponds to the substitution $X = Y', Y = X'$
$$\begin{align}2Y'+3X'&=1\\6Y'-4X'&=2\end{align}$$
Adding $k$ times the second column to the first corresponds to the substitution $Y = Y'+kX$
$$\begin{align}2X+3(Y' + kX)&=(2+3k)X+3Y'&=1\\6X-4(Y' + kX)&=(6-4k)X-4Y'&=2\end{align}$$

So to solve this equation

Multiply the first column by $1/2$: $X = X'/2$
$$\begin{align}X'+3Y&=1\\3X'-4Y&=2\end{align}$$
Add -3 times the first column to the second column: $X' = X''-3Y$
$$\begin{align}X''& &=1\\3X''&-13Y&=2\end{align}$$
Multiply the second column by $-1/13$: $Y=-Y'/13$
$$\begin{align}X''& &=1\\3X''&+Y'&=2\end{align}$$
Add -3 times the second column to the first column: $Y' = Y''-3X''$
$$\begin{align}X''&=1\\Y''&=2\end{align}$$

Now as has been noted, $(1,2)$ is not a solution to the original equation. We need to run all our substitutions backwards to re-express it in terms of $X$ and $Y$.
4. $Y' = Y''-3X''=2-3=-1$
3. $Y = -Y'/13=1/13$
2. $X' = X''-3Y =1-3/13=10/13$
1. $X = X'/2 = (10/13)/2 = 5/13$
Giving the correct solution $X=5/13, Y=1/13$.
This also shows why people prefer row operations. You get the answer in terms of the original variables!
Also note that there was no reason to consider augmented matrices when working this way.