Question

If A = `[(1/2, alpha),(0, 1/2)]`, prove that `sum_("k" = 1)^"n" det("A"^"k") = 1/3(1 - 1/4)` 

Answer 1

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`

|A^k| = `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`

∴ S_n = `("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"]`.