Why does adding an extra column to a matrix not change the RREF of the first columns? This resource (2.2.7) points out that deleting a column from a matrix would not change the RREF of the initial undeleted columns.
For example, the RREF form of this $3\times3$ matrix is the $3\times3$ identity matrix.
\begin{bmatrix}1&1&1\\1&0&1\\1&2&3\end{bmatrix}
The RREF form of this matrix (the above with third column deleted) would be the $3\times 3$ identity matrix with the third column deleted, i.e. the RREF form of the first two colums are preserved.
\begin{bmatrix}1&1\\1&0\\1&2\end{bmatrix}
Is there a way to geometrically, or numerically indicate why the RREF form of individual columns are not changed by adding/deleting other columns? Intuitively, I believe this has something to do with the fact that adding another column to a matrix doesn't change the linear independence of the other columns with respect to each other. Furthermore, how does this connect to deleting or adding rows? Would messing with rows change the RREF form of columns?
 A: Welcome to MSE!
When you bring a matrix into RREF the pivots correspond to the linearly independent columns of your original matrix. See here for how to determine bases of the column space, it's not super crazy. You can translate this question into a basis problem about the column space of the matrix.
Here are two examples that might be of interest to you, you might not get the exact RREF by deleting a column, I chose to work with the $4\times 4$ matrix given from wikipedia:
$$
    \begin{pmatrix}
    1 & 3 & 1 & 4\\
    2 & 7 & 3 & 9\\
    1 & 5 & 3 & 1\\
    1 & 2 & 0 & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0 & -2 & 0\\
    0 & 1 & 1 & 0\\
    0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0\\
    \end{pmatrix}
$$
If you delete the linear dependent column
$\begin{pmatrix}
    1 & 3 & 3 & 0 \\
    \end{pmatrix}^{T} = 
(-2)\cdot\begin{pmatrix}
    1 & 2 & 1 & 1 \\
    \end{pmatrix}^{T}+ 
\begin{pmatrix}
    3 & 7 & 5 & 2 \\
    \end{pmatrix}^{T} $ then
$$
    \begin{pmatrix}
    1 & 3  & 4\\
    2 & 7  & 9\\
    1 & 5  & 1\\
    1 & 2  & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix}
$$
has the same RREF as if you had just deleted the third column. But say I deleted the second column instead, then
$$
    \begin{pmatrix}
    1 & 1  & 4\\
    2 & 3  & 9\\
    1 & 3  & 1\\
    1 & 0  & 8\\
    \end{pmatrix} \rightsquigarrow_{RREF}
\begin{pmatrix}
    1 & 0  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix} \text{not}
\begin{pmatrix}
    1 & -2  & 0\\
    0 & 1  & 0\\
    0 & 0  & 1\\
    0 & 0  & 0\\
    \end{pmatrix}
$$
Deleting linear dependent columns doesn't affect the linear dependence/independence of the remaining columns.
In your example all of the columns were linearly independent (given that you get the RREF is the identity $I_{3\times 3}$), from a basis point of view, removing a vector from a linear independent set does not affect the linear independence of the remaining vectors. Also, the last column was deleted. Having the same RREF with a column deleted 'should' work if all the column vectors are linearly independent.
What I mean by 'should' is, if you deleted the second column instead of the third, your RREF would be
$$
\begin{pmatrix}
    1   & 0\\
    0   & 1\\
    0   & 0\\
    \end{pmatrix} \text{not}
\begin{pmatrix}
    1   & 0\\
    0   & 0\\
    0   & 1\\
    \end{pmatrix}
$$
It wouldn't be exactly the same as just deleting the second column, however we still get the information conveyed that the two column vectors are linearly independent.
If you add a column and performed the same row operations you did to get the RREF of the matrix minus the column, then you would get the same RREF with the column added.
For rows, you can always choose to work with the transpose of the matrix, and then analyze column-wise.
Hope this helps!
