If \(A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\) and n ∈ N (where N is the set of natural numbers), then Aⁿ is equal to

2^n-1A

\(A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]⇒A^2\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 2 & 2 \\ 2 & 2 \end{array}\right]=2A\)

$A^3=A^2.A⇒2A.A=2(A^2)=2.2a=2^{(3-1)}A$

Then in general $A^n=2^{n-1}A$