Final answer:
To determine which of the options holds for all n ∈ N, we can use the properties of matrix exponentiation. The correct option is D. A^n=2^(n-1)A+(n-1)I.
Step-by-step explanation:
To determine which of the options holds for all n ∈ N, we can use the properties of matrix exponentiation. Let's evaluate A^n and compare it with the options given.
A = [1 0] and I = [1 0; 1 1].
A^n = I.A^{n-1} = I.I...(n times) = I^n.A^0 = I^n.I = I^{n+1}
So, the correct option is D. A^n=2^(n-1)A+(n-1)I.