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Question Number 32878 by 7991 last updated on 05/Apr/18

A= [(α,1,1,1),(1,α,1,1),(1,1,β,1),(1,1,1,β) ]with α^2 ≠1≠β^2 det(A)=....???

A=[α1111α1111β1111β]withα2≠1≠β2det(A)=....???

Answered by MJS last updated on 05/Apr/18

det(A)=αdet [(α,1,1),(1,β,1),(1,1,β) ]−det [(1,1,1),(1,β,1),(1,1,β) ]+det [(1,α,1),(1,1,1),(1,1,β) ]−det [(1,α,1),(1,1,β),(1,1,1) ]= [the last 2 are the same, so they sum up to 0] =((αβ^2 +1+1)−(α+β+β))−((β^2 +1+1)−(1+β+β))= =−α^2 +α^2 β^2 −β^2 +4(α−αβ+β)−3= [tried a while, found this:] =(α−1)(β−1)(α+αβ+β−3)

det(A)=αdet[α111β111β]−det[1111β111β]+det[1α111111β]−det[1α111β111]=[thelast2arethesame,sotheysumupto0]=((αβ2+1+1)−(α+β+β))−((β2+1+1)−(1+β+β))==−α2+α2β2−β2+4(α−αβ+β)−3=[triedawhile,foundthis:]=(α−1)(β−1)(α+αβ+β−3)

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