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If `omega!=1` is a complex cube root of unity, then prove that `[{:(1+2omega^(2017)+omega^(2018)," "omega^(2018),1),(1,1+2omega^(2018)+omega^(2017),omega^(2017)),(omega^(2017),omega^(2018),2+2omega^(2017)+omega^(2018)):}]`is singular |
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Answer» Let `A=[{:(1+2omega^(2017)+omega^(2018)," "omega^(2018),1),(1,1+2omega^(2018)+omega^(2017),omega^(17)),(omega^(17),omega^(18),2+2omega^(2017)+omega^(2018)):}]` ` therefore" " omega^(3)=1rArr omega^(2017)=omega` and `omega^(2018)=omega^(2)` then `[(1+2omega+omega^(2),omega^(2),1),(1,1+omega^(2)+2omega,omega),(omega,omega^(2),2+omega+2omega^(2))]` `=[(omega,omega^(2),1),(1,omega,omega),(omega,omega^(2),-omega)]" " [therefore1+omega+omega^(2)=0]` Now, `|A|= [(omega,omega^(2),1),(1,omega,omega),(omega,omega^(2),-omega)]= omega[(omega,omega,1),(1,1,omega),(omega,omega,-omega)]=0 thus, `|A|=0.` Hence, A is singular matrix. |
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