If A is an idempotent matrix and I is an identify matrix of the Same order, then the value of n, such that `(A+I)^n =I+127A` is

Answer

7

Explanation

`becauseA` is idempotent matrix
`therefore A^(2)=A`
` rArr A =A^(2) =A^(3) = A^(4) = A^(5) = ... " " …(i)`
Now, `(A+I) ^(n) = (I + A)^(n)`
` = I+ ""^(n)C_(1) + ""^(n) C_(2) A^(2) + ""^(n) C_(3) A^(3) + ... + ""^(n) C_(n) A^(n) `
` = I+ (""^(n)C_(1) + ""^(n) C_(2) + ""^(n) C_(3) + ... + ""^(n) C_(n) )A`
[from Eq. (i)]
`rArr (A+I) ^(n) = I + (2^(n) - 1) A " " ...(ii)`
Given, we get
`(A + I)^(n) = I + 127A " " ...(iii)`
From Eqs (ii) and (iii), we get
`2^(n) - 1 = 127`
`rArr 2^(n) = 128 = 2^(7) `
`rArr therefore n= 7`