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)=....???](https://www.tinkutara.com/question/Q32878.png)
$${A}=\begin{bmatrix}{\alpha}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\alpha}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}{with}\:\alpha^{\mathrm{2}} \neq\mathrm{1}\neq\beta^{\mathrm{2}} \\ $$$${det}\left({A}\right)=….??? \\ $$
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)](https://www.tinkutara.com/question/Q32884.png)
$$\mathrm{det}\left({A}\right)=\alpha\mathrm{det}\begin{bmatrix}{\alpha}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\beta}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}+\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\end{bmatrix}−\mathrm{det}\begin{bmatrix}{\mathrm{1}}&{\alpha}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\beta}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{bmatrix}= \\ $$$$\:\:\:\:\:\left[\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same},\:\mathrm{so}\:\mathrm{they}\right. \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{sum}\:\mathrm{up}\:\mathrm{to}\:\mathrm{0}\right] \\ $$$$=\left(\left(\alpha\beta^{\mathrm{2}} +\mathrm{1}+\mathrm{1}\right)−\left(\alpha+\beta+\beta\right)\right)−\left(\left(\beta^{\mathrm{2}} +\mathrm{1}+\mathrm{1}\right)−\left(\mathrm{1}+\beta+\beta\right)\right)= \\ $$$$=−\alpha^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} −\beta^{\mathrm{2}} +\mathrm{4}\left(\alpha−\alpha\beta+\beta\right)−\mathrm{3}= \\ $$$$\:\:\:\:\:\left[\mathrm{tried}\:\mathrm{a}\:\mathrm{while},\:\mathrm{found}\:\mathrm{this}:\right] \\ $$$$=\left(\alpha−\mathrm{1}\right)\left(\beta−\mathrm{1}\right)\left(\alpha+\alpha\beta+\beta−\mathrm{3}\right) \\ $$