# Characteristic polynomial calculation method

$$\begin{vmatrix}3-\lambda & -4 & -2 & -4\\ -2 & 5-\lambda & 2 & 4\\ 4 & -8 & -3-\lambda & -8\\ 2 & -4 & -2 & -3-\lambda \end{vmatrix}=\begin{vmatrix}3 & -4 & -2 & -4\\ -2 & 5 & 2 & 4\\ 4 & -8 & -3 & -8\\ 2 & -4 & -2 & -3 \end{vmatrix}+\lambda(\begin{vmatrix}5 & 2 & 4\\ -8 & -3 & -8\\ -4 & -2 & -3 \end{vmatrix}+\begin{vmatrix}3 & -2 & -4\\ 4 & -3 & -8\\ 2 & -2 & -3 \end{vmatrix}+\begin{vmatrix}3 & -4 & -4\\ -2 & 5 & 4\\ 2 & -4 & -3 \end{vmatrix}+\begin{vmatrix}3 & -4 & -2\\ -2 & 5 & 2\\ 4 & -8 & -3 \end{vmatrix}) + \lambda^{2}(\begin{vmatrix}\\ \\ \end{vmatrix}+\begin{vmatrix}\\ \\ \end{vmatrix}+\begin{vmatrix}\\ \\ \end{vmatrix}+\begin{vmatrix}\\ \\ \end{vmatrix})+\lambda^{3}(3+5-3-3)+\lambda^{4}=$$

Thanks in avance for help

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You're going to need 6 matrices for that $\lambda^2$ term. Also, I think that should be a minus sign out in front of $\lambda^3$. – Gerry Myerson Feb 27 '13 at 7:39
Also, please check the signs of the $\lambda$ and $\lambda^{3}$ coefficients. – Andreas Caranti Feb 27 '13 at 7:41
@GerryMyerson Could You tell me more, how to calculate them? – Steve Feb 27 '13 at 7:49
The easiest method for me is sage – Lior B-S Feb 27 '13 at 8:43
With apologies for the formatting, they would be [3, -4, -2, 5], [3, -2, 4, -3], [3, -4, 2, -3], [5, 2, -8, -3], [5, 4, -4, -3], and [-3, -8, -2, -3]. – Gerry Myerson Feb 27 '13 at 12:16

$\def\p{\phantom{-}}$
Upgrading comment to answer, at request of OP: $$\pmatrix{\p3&-4\cr-2&\p5\cr},\pmatrix{3&-2\cr4&-3\cr},\pmatrix{3&-4\cr2&-3\cr},\pmatrix{\p5&\p2\cr-8&-3\cr},\pmatrix{\p5&\p4\cr-4&-3\cr},\pmatrix{-3&-8\cr-2&-3\cr}$$