If `[cos(2pi)/7-sin(2pi)/7sin(2pi)/7cos(2pi)/7]=[1 0 0 1]` , then the least positive integral value of `k` is (a) 3 (b) 4 (c) 6 (d) 7

Options

A. 3
B. 4
C. 6
D. 7

Answer

D

Explanation

We have,
`{:[(cos.(2pi)/7,-sin.(2pi)/7),(sin.(2pi)/7,cos.(2pi)/7)]^2:}`
`{:=[(cos.(2pi)/7,-sin.(2pi)/7),(sin.(2pi)/7,cos.(2pi)/7)][(cos.(2pi)/7,-sin.(2pi)/7),(sin.(2pi)/7,cos.(2pi)/7)]:}`
`{:=[(cos^2.(2pi)/7-sin.(2pi)/7,-2sin.(2pi)/7cos.(2pi)/7),(2sin.(2pi)/7cos.(2pi)/7,-sin^2.(2pi)/7+cos^2.(2pi)/7)]:}`
`{:=[(cos.(4pi)/7,-sin.(4pi)/7),(sin.(4pi)/7,cos.(4pi)/7)]:}`
Continuing in this manner, we get
`{:[(cos.(2pi)/7,-sin.(2pi)/7),(sin.(2pi)/7,cos.(2pi)/7)]^7=[(cos2pi,-sin2pi),(sin2pi,cos2pi)]=[(0,1),(0,1)]:}`
Hence, the least value of n is 7.