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Question
If A = `[(1/2, alpha),(0, 1/2)]`, prove that `sum_("k" = 1)^"n" det("A"^"k") = 1/3(1 - 1/4)`
Solution
A = `[(1/2, alpha),(0, 1/2)]`
|A| = `|(1/2,alpha),(0, 1/2)|`
= `1/4 - 0`
= `1/4`
A2 = A × A
= `[(1/2, alpha),(0, 1/2)] [(1/2, alpha),(0, 1/2)]`
= `[(1/4, alpha),(0, 1/4)]`
|A2| = `|(1/4, alpha),(0, 1/4)|`
=`1/4 xx 1/4 - 0`
= `(1/4)^2`
= `1/4^2`
|Ak| = `1/4^"k"`
So `sum_("k" = 1)^"n" det("A"^"k") = 1/4 + 1/4^2 + 1/4^3 + ...... + 1/4^"n"`
Which is a G.P with a `1/4` and r = `1/4`
∴ Sn = `("a"(1 - "r"^"n"))/(1 - "r")`
= `(1/4[1 - (1/4)^"n"])/(1 - 1/4)`
= `(1/4[1 - 1/4^"n"])/(3/4)`
= `1/4 xx 4/3[1 - 1/4^"n"]`
= `1/3[1 - 1/4^"n"]`.
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