From 5391c0f3e5a23b4e8638b8ec7b70677907438fc3 Mon Sep 17 00:00:00 2001 From: Aaron Huber Date: Wed, 4 Nov 2020 16:31:39 -0500 Subject: [PATCH] Fixes to 2.6 in computing determinant. --- poly-form.tex | 24 ++++++++++++------------ 1 file changed, 12 insertions(+), 12 deletions(-) diff --git a/poly-form.tex b/poly-form.tex index dd5a84b..0b7744c 100644 --- a/poly-form.tex +++ b/poly-form.tex @@ -587,8 +587,8 @@ For the LHS we get We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. To make it easier, use the following variable representations: $x = \numocc{\graph{1}}{\tri}, y = \numocc{\graph{1}}{\threepath}, z = \numocc{\graph{1}}{\threedis}$. Using $\linsys{2}$ and $\linsys{3}$, the following matrix is obtained, \[ \mtrix{\rpoly} = \begin{pmatrix} 1 & \prob & -(3\prob^2 - \prob^3)\\ --2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\ --18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3) +-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\ +-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3) \end{pmatrix},\] and the following linear equation \begin{equation} @@ -607,18 +607,18 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a \begin{equation*} \begin{vmatrix} 1 & \prob & -(3\prob^2 - \prob^3)\\ --2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\ --18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3) +-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & 10(3\prob^2 - \prob^3)\\ +-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & 45(3\prob^2 - \prob^3) \end{vmatrix} = (3\prob^2 - \prob^3)^2 \cdot \begin{vmatrix} --4 & -2\\ --21 & -3 +-4 & 10\\ +-21 & 45 \end{vmatrix} ~ - ~ \prob(3\prob^2 - \prob^3)^2~ \cdot \begin{vmatrix} --2 & -2\\ --18 & -3 +-2 & 10\\ +-18 & 45 \end{vmatrix} - ~(3\prob^2 - \prob^3)^3~ \cdot \begin{vmatrix} @@ -629,11 +629,11 @@ We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $a Compute each RHS term starting with the left and working to the right, \begin{equation} -(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot -3) - (-21 \cdot -2)\right) = (3\prob^2 - \prob^3)^2\cdot(12 - 42) = -30(3\prob^2 - \prob^3)^2.\label{eq:det-1} +(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot 45) - (-21 \cdot 10)\right) = (3\prob^2 - \prob^3)^2\cdot(-180 + 210) = 30(3\prob^2 - \prob^3)^2.\label{eq:det-1} \end{equation} The middle term then is \begin{equation} --\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot -3) - (-18 \cdot -2)\right) = -\prob(3\prob^2 - \prob^3)^2 \cdot ( 6 - 36) = 30\prob(3\prob^2 - \prob^3)^2.\label{eq:det-2} +-\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot 45) - (-18 \cdot 10)\right) = -\prob(3\prob^2 - \prob^3)^2 \cdot (-90 + 180) = -90\prob(3\prob^2 - \prob^3)^2.\label{eq:det-2} \end{equation} Finally, the rightmost term, \begin{equation} @@ -642,8 +642,8 @@ Finally, the rightmost term, Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have, \begin{align} -\dtrm{\mtrix{\rpoly}} =& -30(3\prob^2 - \prob^3)^2 + 30\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3 = 30(3\prob^2 - \prob^3)^2\left(-1 + \prob + (3\prob^2 - \prob^3)\right) = 30\left(9\prob^4 - 6\prob^5 + \prob^6\right)\left(-p^3 + 3p^2 + p - 1\right)\nonumber\\ -=&\left(30\prob^6 - 180\prob^5 + 270\prob^4\right)\cdot\left(-p^3 + 3p^2 + p - 1\right).\label{eq:det-final} +\dtrm{\mtrix{\rpoly}} =& 30(3\prob^2 - \prob^3)^2 - 90\prob(3\prob^2 - \prob^3)^2 +30(3\prob^2 - \prob^3)^3 = 30(3\prob^2 - \prob^3)^2\left(1 - 3\prob + (3\prob^2 - \prob^3)\right) = 30\left(9\prob^4 - 6\prob^5 + \prob^6\right)\left(-\prob^3 - 3\prob^2 + 3\prob + 1\right)\nonumber\\ +=&\left(30\prob^6 - 180\prob^5 + 270\prob^4\right)\cdot\left(-\prob^3 - 3\prob^2 + 3\prob + 1\right).\label{eq:det-final} \end{align} \AH{It appears that the equation below has roots at p = 0 (left factor) and p = 0.4608 (right factor), \textit{UNLESS} I made a mistake somewhere.}