1. We are given

$$P=\begin{bmatrix}1&0\\ \tfrac12 &1\end{bmatrix}.$$

We need to find $P^{50}$.

2. Write $P$ as

$$P=I+N,$$

where

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad N=\begin{bmatrix}0&0\\ \tfrac12 &0\end{bmatrix}.$$

3. Now compute $N^2$:

$$N^2= \begin{bmatrix}0&0\\ \tfrac12 &0\end{bmatrix} \begin{bmatrix}0&0\\ \tfrac12 &0\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}.$$

So, $N$ is nilpotent of order $2$.

4. Therefore,

$$P^{50}=(I+N)^{50}.$$

Using the binomial theorem for matrices (valid here since $I$ and $N$ commute),

$$(I+N)^{50}=I+50N$$

because all higher powers vanish as $N^2=0$.

5. Now compute:

$$50N=50\begin{bmatrix}0&0\\ \tfrac12 &0\end{bmatrix} = \begin{bmatrix}0&0\\25&0\end{bmatrix}.$$

Hence,

$$P^{50}=I+50N= \begin{bmatrix}1&0\\25&1\end{bmatrix}.$$

6. Comparing with the options:

• A: $\begin{bmatrix}1&0\\25&1\end{bmatrix}$ ✅
• B: $\begin{bmatrix}1&50\\0&1\end{bmatrix}$ ❌
• C: $\begin{bmatrix}1&25\\0&1\end{bmatrix}$ ❌
• D: $\begin{bmatrix}1&0\\50&1\end{bmatrix}$ ❌

Therefore, the correct option is A.