# Intuiting Product of Elimination Matrices (and NOT by Matrix Multiplication)

I want to intuit, and NOT compute with matrix multiplication, $M:=\color{green}{E_{P_3 \rightarrow P_4}}\color{#CA790F}{E_{P_2 \rightarrow P_3}}E_{P_1 \rightarrow P_2},$ where:

$E_{P_1 \rightarrow P_2} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ \end{bmatrix}, \color{#CA790F}{E_{P_2 \rightarrow P_3} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ \end{bmatrix}}, \color{green}{E_{P_3 \rightarrow P_4} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ \end{bmatrix}}$.

$\fbox{1st row :}$ None of these elimination matrices affect the first row.
So the first row of $M = (1, 0, 0,0)$.

$\fbox{2nd row :}$ By $E_{P_1 \rightarrow P_2}$, $-R_1 + R_2 \rightarrow R_2$. So the 2nd row of $M = (-1, 1, 0,0)$

$\fbox{3rd row :}$ By $E_{P_1 \rightarrow P_2}, -R_2 + R_3 \rightarrow R_3^{*}.$
By $\color{#CA790F}{E_{P_2 \rightarrow P_3}}$, $-R_2 + R_3^* \rightarrow R_3^{**}$.
So the 3rd row of $M = (0, -1\color{#CA790F}{-1}, 1,0)$.
Here, the third column is $1$ and NOT $2$ because $R_3$ is not being added to itself twice. It only needs to appear once.

$\fbox{4th row :}$ By $E_{P_1 \rightarrow P_2}$, $-R_3 + R_4 \rightarrow R_4^*$.
By $\color{#CA790F}{E_{P_2 \rightarrow P_3}}$, $-R_3 + R_4^* \rightarrow R_4^{**}$.
By $\color{green}{E_{P_3 \rightarrow P_4}}$, $-R_3 + R_4^{**} \rightarrow R_4^{***}$.
So the 4th row of $M = (0, 0, -1\color{#CA790F}{-1}\color{green}{-1},1)$. Here, the 4th column is $1$ and NOT $3$ because $R_4$ is not being added to itself thrice. It is $-R_3$ which is added (to $R_4$) thrice.

Altogether, my $M = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -2 & \color{#668FE2}{1} & 0 \\ 0 & 0 & -3 & \color{#668FE2}{1} \\ \end{bmatrix}.$ However, the correct $M = \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ \color{red}{1} & -2 & \color{#668FE2}{1} & 0 \\ \color{red}{-1} & \color{red}{3} & -3 & \color{#668FE2}{1} \\ \end{bmatrix}.$
How and why did I miss the three red entries?
Also, is there a better explanation why the $\color{#668FE2}{1}$s are not $2$ in the 3rd row and $3$ in the 4th row?

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I'm sort of guessing at the problem, but it's not correct to compute $M$ "row-by-row". You ought to start with $4 \times 4$ identity and apply all of the row operations corresponding to $E_{P_1 \rightarrow P_2}$, then all of the row operations corresponding to $E_{P_2 \rightarrow P_3}$ and finally the ones corresponding to $E_{P_3 \rightarrow P_4}$, in that order. Does that make sense? As opposed to applying the 1st row operation of each of each matrix, then applying the 2nd row operation of each matrix, and so on. – Mike F Aug 18 '13 at 3:31