Let $$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3 & 3 & 1 \end{bmatrix} $$
Which, I want to convert to $I_{4\times4}$ by premultiplying by elimination matrices.
Now, the row operations are:
First column: $R_2 \leftarrow R_2 - R_1, R_3 \leftarrow R_3 - R_1, R_4 \leftarrow R_4 - R_1$
Second column: $R_2 \leftarrow R_3 - 2R_2, R_4 \leftarrow R_4 - 3R_2$
Third column: $R_4 \leftarrow R_4 - 3R_3$
Now, when I eliminate one column at a time, I get $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -2 & 1 & 0 \\ 0 & -3 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \end{bmatrix} A = I_{4\times4} $$ Where each matrix eliminates entries from each successive column. When I multiply those matrices together, I get $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ -1 & 3 & -3 & 1 \end{bmatrix}A = I $$ and thus, this matrix is the inverse of $A$.
However, if I try to write eliminate all columns at the same time, using only one elimination matrix and the row transforms described above, I get $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ -1 & -2 & 1 & 0 \\ -1 & -3 & -3 & 1 \end{bmatrix}A \neq I $$
Why is there such discrepancy? Why does eliminating one column at a time gives the correct answer whereas eliminating all columns simultaneously does not?