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

$$\left|\begin{array}{cccccc}\epsilon & 1& 0& 0& 0& 1\\ 1& \epsilon & 1& 0& 0& 0\\ 0& 1& \epsilon & 1& 0& 0\\ 0& 0& 1& \epsilon & 1& 0\\ 0& 0& 0& 1& \epsilon & 1\\ 1& 0& 0& 0& 1& \epsilon \end{array}\right|=?$$$$$$

Answered by SdC355 last updated on 30/Mar/25

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

Commented by MrGaster last updated on 30/Mar/25

$$\mathcal{THANK}!$$

Commented by Ghisom last updated on 30/Mar/25

$$\mathrm{the}‘‘\mathrm{holy}{\mathrm{shit}}^{″}\mathrm{is}\mathrm{back}.\mathrm{nice}.$$

Answered by Wuji last updated on 30/Mar/25

$$forthe6\times 6matrix$$$$M\left(\epsilon \right)=\epsilon I+A$$$$foracirculantgraph{C}_{n}wehaveeigenvalues$$$$ofA:$$$$2cos\left(\frac{2\pi k}{n}\right)\Rightarrow k=0,1,...,n-1$$$$forn=6$$$$2,1,-1,-2,-1,1$$$$M\left(\epsilon \right)=\epsilon I+A$$$$\epsilon +2,{(\epsilon +1)}^2,{(\epsilon -1)}^2,(\epsilon -2)$$$$detM\left(\epsilon \right)=(\epsilon +2){(\epsilon +1)}^2{(\epsilon -1)}^2(\epsilon -2)$$$$detM\left(\epsilon \right)=({\epsilon }^2-4){({\epsilon }^2-1)}^2$$$$$$

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