If $$A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ and $$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$, then which one of the following holds for all $$n \geq 1$$, by the principle of mathematical induction

Define the matrix $$A = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$ and the identity matrix $$I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$. We have to show, for every integer $$n \ge 1$$, that

$$A^{\,n} = nA - (n-1)I \quad\quad -(1)$$

We prove statement (1) by the Principle of Mathematical Induction (PMI).

Base case ($$n = 1$$):
$$A^{\,1} = A$$ and $$1A - (1-1)I = A - 0\cdot I = A$$.
Thus (1) is true for $$n = 1$$.

Inductive hypothesis:
Assume (1) is true for some $$n = k \; (\ge 1)$$, i.e.

$$A^{\,k} = kA - (k-1)I \quad\quad -(2)$$

Inductive step ($$n = k+1$$):

Start with $$A^{\,k+1} = A^{\,k}\,A$$ and substitute (2):

$$A^{\,k+1} = \bigl(kA - (k-1)I\bigr)A = kA^{\,2} - (k-1)A \quad\quad -(3)$$

First compute $$A^{\,2}$$ once for all:

$$A^{\,2} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix} \quad\quad -(4)$$

Insert (4) into (3):

$$A^{\,k+1} = k\begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix} - (k-1)\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$ $$= \begin{bmatrix}k & 0 \\ 2k & k\end{bmatrix} - \begin{bmatrix}k-1 & 0 \\ k-1 & k-1\end{bmatrix}$$ $$= \begin{bmatrix}k-(k-1) & 0 \\ 2k-(k-1) & k-(k-1)\end{bmatrix}$$ $$= \begin{bmatrix}1 & 0 \\ k+1 & 1\end{bmatrix} \quad\quad -(5)$$

Now evaluate the right-hand side of (1) for $$n = k+1$$:

$$(k+1)A - kI = (k+1)\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix} - k\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ $$= \begin{bmatrix}k+1 & 0 \\ k+1 & k+1\end{bmatrix} - \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}$$ $$= \begin{bmatrix}1 & 0 \\ k+1 & 1\end{bmatrix} \quad\quad -(6)$$

Expressions (5) and (6) are identical, so

$$A^{\,k+1} = (k+1)A - kI$$

Consequently (1) holds for $$n = k+1$$ whenever it holds for $$n = k$$. By the Principle of Mathematical Induction, statement (1) is true for all integers $$n \ge 1$$.

Hence the correct option is:
Option A which is: $$A^{\,n} = nA - (n-1)I$$.