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Question Number 189254 by yaslm last updated on 13/Mar/23

Answered by manxsol last updated on 14/Mar/23

determinant ((a,1,1,1,1,1,1),(1,a,1,1,1,1,1),(,,a,,,,1),(,,,.,,,),(1,1,1,1,a,1,1),(1,1,1,1,1,a,1),(1,1,1,1,1,1,a)) −fn+f_(n−1) −fn+f_(n−2) . −f_n +f1 determinant (((a−1),0,0,0,0,0,(1−a)),(0,(a−1),0,0,0,0,(1−a)),(0,,(a−1),,,,(1−a)),(,,,.,,,),(0,0,0,0,(a−1),0,(1−a)),(0,0,0,0,0,(a−1),(1−a)),(1,1,1,1,1,1,a)) determinant (((a−1),0,0,0,0,0,0),(0,(a−1),0,0,0,0,0),(0,,(a−1),,,,0),(,,,.,,,),(0,0,0,0,(a−1),0,0),(0,0,0,0,0,(a−1),0),(1,1,1,1,1,1,(a+n−1))) c_1 +c_n c_2 +c_n .. c_(n−1) +c triangular superior D=(a−1)^(n−1) (a+n−1) aplication a=3 D=2^(n−1) (2+n)

$$\begin{vmatrix}{{a}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{{a}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{}&{}&{{a}}&{}&{}&{}&{\mathrm{1}}\\{}&{}&{}&{.}&{}&{}&{}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}\end{vmatrix} \\ $$$$−{fn}+{f}_{{n}−\mathrm{1}} \\ $$$$−{fn}+{f}_{{n}−\mathrm{2}} \\ $$$$. \\ $$$$−{f}_{{n}} +{f}\mathrm{1} \\ $$$$ \\ $$$$\begin{vmatrix}{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}−{a}}\\{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}−{a}}\\{\mathrm{0}}&{}&{{a}−\mathrm{1}}&{}&{}&{}&{\mathrm{1}−{a}}\\{}&{}&{}&{.}&{}&{}&{}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}−{a}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{1}−{a}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}}\end{vmatrix} \\ $$$$\begin{vmatrix}{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{}&{{a}−\mathrm{1}}&{}&{}&{}&{\mathrm{0}}\\{}&{}&{}&{.}&{}&{}&{}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{{a}−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{{a}+{n}−\mathrm{1}}\end{vmatrix} \\ $$$${c}_{\mathrm{1}} +{c}_{{n}} \\ $$$${c}_{\mathrm{2}} +{c}_{{n}} \\ $$$$.. \\ $$$${c}_{{n}−\mathrm{1}} +{c} \\ $$$${triangular}\:{superior} \\ $$$${D}=\left({a}−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({a}+{n}−\mathrm{1}\right) \\ $$$${aplication} \\ $$$${a}=\mathrm{3} \\ $$$${D}=\mathrm{2}^{{n}−\mathrm{1}} \left(\mathrm{2}+{n}\right) \\ $$

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