If `A=[(1,a),(0, 1)]` , then `A^n` (where `n in N)` equals (a)`[(1,n a),(0, 1)]` (b) `[(1,n^2a),(0, 1)]` (c) `[(1,n a),(0 ,0)]` (d) `[(n,n a),(0,n)]`

Options

A. `{:A=[(1,na),(0,1)]:}`
B. `{:A=[(1,n^2a),(0,1)]:}`
C. `{:A=[(1,na),(0,0)]:}`
D. `{:A=[(1,2a),(0,n)]:}`

Answer

B

Explanation

We have, `{:A=[(1,a),(0,1)]:}`
`:.A^2={:[(1,a),(0,1)][(1,a),(0,1)]=[(1,2a),(0,1)]:}`
It can proved by the principle of mathematical induction that
`A^n={:[(1,na),(0,1)]:}` for all `n in N.