Mr Real said:
Oh, I think this is the source of much of the confusion, because I didn't ever say or imply that all three matrices are subject to the same operation, I only talked about applying the operation in the RHS as well as the LHS, but since there are two matrices in the LHS, maybe you thought I was talking about changing both of them, when in fact I was talking only about one.
In particular, you need to talk about applying the row operation only the
first factor in a RHS that is a product of matrices ( rather than being something like FG + H)
Now can you tell me why for an equality; C = (F)(G), some operations, like changing R2 to R2+R3 work while some other operations like changing R2 to R1+R3 don't, in context of linear transformations?(I have understood the reasons for this when we take matrices to be a system of equations)
Again, we must define what it means for something to "work".
For example, the linear transformation given by:
##x_{new} = 2 x_{old} + 3 y_{old},##
##y_{new} = x_{old} - y_{old}##
defines a transformation (i.e. a function) that maps coordinates ##(x_{old},y_{old})## to coordinates ##(x_{new},y_{new})##. There no overt notion of
solving for anything in the definition of a linear transformation.
However, we can think a "solutions" to the linear transformation as being quadruples of numbers of the form ##(x_{new},y_{new},x_{old},y_{old})## (such as (5,0, 1,1) in the example) where the second pair of numbers is the result of applying the linear transformation to the first pair of numbers. From this point of view, not all quadruples of numbers are solutions. For example (1,2,3,4) is not a solution to the example.
Taking that point of view, two linear transformations are "equivalent" when they have the same set of solutions.
Since your question involves 3 rows, we must look at an example of a linear transformation like:
##\begin{pmatrix} x_{new}\\ y_{new}\\ z_{new} \end{pmatrix} = \begin{pmatrix}3&1&1\\0&1&0 \\2&0&5\end{pmatrix} \begin{pmatrix} x_{old}\\y_{old}\\z_{old} \end{pmatrix} ##
This matrix equation can be interpreted as 3 linear equations (in 6 variables) that define the linear transformation:
eq. 1) ##x_{new} = 3x_{old} + y_{old} + z_{old}##
eq. 2) ##y_{new} = y_{old}##
eq. 3) ##z_{new} = 2x_{old} + 5 z_{old}##
(When people say that a matrix "is" a linear transformation, they mean that you can supply variables to form a matrix equation like the one above.)The customary way to write a system of equations is to put all the variables on the left hand side, but we don't need to do that for our purposes.
We have 3 equations in 6 unknowns, so we expect the system to have an infinite number of solutions. However, having an infinite number of solutions does not imply that any arbitrary sextuple of numbers will be one of those solutions.
Suppose we do the operation "Replace row 2 by the sum of row 1 and row 3". That's like erasing eq 2.) and putting the sum of equation 1) and equation 3) in its place. Would you expect that operation to preserve the solutions to the system of equations? It removes all information about ##y_{new}## and replaces it with an equation that is just a consequence of eq. 1) and eq. 3).
