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. 6
C. 7
D. 14

Answer

C

Explanation

We know that `[(cos theta,- sin theta),(sin theta,cos theta)]^(n)=[(cos n theta,-sin n theta),(sin n theta,cos n theta)]`
`:. [("cos" (2pi)/7,-"sin" (2pi)/7),("sin"(2pi)/7,"cos"(2pi)/7)]^(k)=[("cos"(2kpi)/7,-"sin"(2kpi)/7),("sin" (2kpi)/7,"cos" (2kpi)/7)]`
`=[(1,0),(0,1)]`, if `k=7m`, where `m in N`
Hence, the least value of k is 7.