Saturday 17 August 2013

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

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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