If A is a square matrix such that $A^2 = A$ and $(I+A)^n = I +λA$, then λ =

$2^n-1$

We have,

$A^2 = A$

$∴(I+A)^2 = (I + A) (I+A) = I + 2A + A^2 = I + 3A$

and, $(I + A)^3 = (I + A)^2 (I + A)$

$= (I + 3A) (I + A)$  $[∵ (I+A)^2 = I + 3A]$

$= I + 4A + 3A^2 = I +7A$  $[∵ A^2 = A]$

Thus, we have

$(I + A)^2 = I + 3A$ and $(I + A)^3 = I +7A$

$⇒(I + A)^2 = I + (2^2 −1) A$ and $(I + A)^3 = I + (2^3-1) A$

Hence, $(I+A)^n = I +(2^n -1) A$

$∴(I+A)^n =I+λA⇒ λ=2^n-1$