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Question Number 218119 by MrGaster last updated on 30/Mar/25

determinant ((ε,1,0,0,0,1),(1,ε,1,0,0,0),(0,1,ε,1,0,0),(0,0,1,ε,1,0),(0,0,0,1,ε,1),(1,0,0,0,1,ε))=?

$$\begin{vmatrix}{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}\end{vmatrix}=? \\ $$$$ \\ $$

Answered by SdC355 last updated on 30/Mar/25

Holy shit....what is that Lol But that det{A} is ε^6 −6ε^4 +9ε^2 −4...

$$\mathrm{Holy}\:\mathrm{shit}....\mathrm{what}\:\mathrm{is}\:\mathrm{that}\:\mathrm{Lol} \\ $$$$\mathrm{But}\:\mathrm{that}\:\mathrm{det}\left\{\mathrm{A}\right\}\:\mathrm{is}\:\varepsilon^{\mathrm{6}} −\mathrm{6}\varepsilon^{\mathrm{4}} +\mathrm{9}\varepsilon^{\mathrm{2}} −\mathrm{4}... \\ $$

Commented by MrGaster last updated on 30/Mar/25

THANK!

$$\mathscr{THANK}! \\ $$

Commented by Ghisom last updated on 30/Mar/25

the “holy shit” is back. nice.

$$\mathrm{the}\:``\mathrm{holy}\:\mathrm{shit}''\:\mathrm{is}\:\mathrm{back}.\:\mathrm{nice}. \\ $$

Answered by Wuji last updated on 30/Mar/25

for the 6×6 matrix M(ε)=εI+A for a circulant graph C_n we have eigenvalues of A : 2cos(((2πk)/n)) ⇒k=0,1,...,n−1 for n=6 2,1,−1,−2,−1,1 M(ε)=εI + A ε+2 ,(ε+1)^2 ,(ε−1)^2 ,(ε−2) detM(ε)=(ε+2)(ε+1)^2 (ε−1)^2 (ε−2) detM(ε)=(ε^2 −4)(ε^2 −1)^2

$${for}\:{the}\:\mathrm{6}×\mathrm{6}\:{matrix} \\ $$$${M}\left(\varepsilon\right)=\varepsilon{I}+{A} \\ $$$${for}\:{a}\:{circulant}\:{graph}\:{C}_{{n}} \:{we}\:{have}\:{eigenvalues} \\ $$$${of}\:{A}\:: \\ $$$$\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi{k}}{{n}}\right)\:\:\:\Rightarrow{k}=\mathrm{0},\mathrm{1},...,{n}−\mathrm{1}\: \\ $$$${for}\:{n}=\mathrm{6}\:\: \\ $$$$\mathrm{2},\mathrm{1},−\mathrm{1},−\mathrm{2},−\mathrm{1},\mathrm{1} \\ $$$${M}\left(\varepsilon\right)=\varepsilon{I}\:+\:{A} \\ $$$$\varepsilon+\mathrm{2}\:,\left(\varepsilon+\mathrm{1}\right)^{\mathrm{2}} ,\left(\varepsilon−\mathrm{1}\right)^{\mathrm{2}} ,\left(\varepsilon−\mathrm{2}\right) \\ $$$${detM}\left(\varepsilon\right)=\left(\varepsilon+\mathrm{2}\right)\left(\varepsilon+\mathrm{1}\right)^{\mathrm{2}} \left(\varepsilon−\mathrm{1}\right)^{\mathrm{2}} \left(\varepsilon−\mathrm{2}\right) \\ $$$${detM}\left(\varepsilon\right)=\left(\varepsilon^{\mathrm{2}} −\mathrm{4}\right)\left(\varepsilon^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$$ \\ $$

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