If `A = [[1 ,1],[1,1]]` and det `(A^(n) - 1) = 1 -lambda ^(n), n in N,` then the value of `lambda` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

`because " "A = [[1 ,1],[1,1]]`
`therefore " "A^(2) = [[2 ,2],[2,2]] = 2[[1 ,1],[1,1]] =2A`
` rArr A^(3) = A^(2) cdot A = 2A^(2) = 2^(2) A`
Similarly, `A_(n) = 2^(n-1) A `
`therefore " "A^(n) -I= [[2^(n-1) ,2^(n-1)],[2^(n-1),2^(n-1)]] -[[1 ,0],[0,1]]`
`= [[2^(n-1)-1 ,2^(n-1)],[2^(n-1),2^(n-1)-1]] `
`rArr "det" (A^(n)-I) = (2^(n-1)-1)^(2) - (2^(n-1))^(2)`
`= 1- 2^(n) = 1 - lambda ^(n)` [given]
`therefore lambda = 2`