Linear Algebra: Finding a steady state matrix Here is the problem:

And here is what I tried to do:

I tried doing it in my calculator and got that the answer is 40 and 160 as you keep multiplying PX. Can anyone point out what I'm doing wrong or help me?
Thanks.
Edit for readability:
Problem Statement:
A college dormitory houses 200 students. Those who watch an hour or more of TV on any day always watch for less than an hour the next day. One-fourth of those who watch TV for less than an hour one day will watch an hour or more the next day. Half of the students watch TV for an hour or more today.
a) How many will watch TV for an hour or more tomorrow?
b) In 2 days?
c) Find the steady state matrix for populations described in this problem.
Work:
$
P = \stackrel{>1 \hphantom{XX} <1}{\begin{bmatrix} 
0 & 0.25\\ 
1 & 0.75 
\end{bmatrix}} {>1 \atop <1}
\quad
x = \begin{bmatrix}
100\\
100
\end{bmatrix}
{>1 \atop <1}
$
$
Px = \begin{bmatrix}
0 & 0.25\\
1 & 0.75
\end{bmatrix}
\begin{bmatrix}
100\\
100
\end{bmatrix}
=
\begin{bmatrix}
25\\
175
\end{bmatrix}
{>1 \atop }
$
a) 25.
$
P^{\,2}x = \begin{bmatrix}
0 & 0.25\\
1 & 0.75
\end{bmatrix}
\begin{bmatrix}
25\\
175
\end{bmatrix}
=
\begin{bmatrix}
43.75\\
156.25
\end{bmatrix}
{>1 \atop <1}
$
b) $ 43.75 \sim 44 $.
Steady state $P\,\overline{x} = \overline{x}$
\begin{bmatrix}
0 & 0.25\\
1 & 0.75
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}
\end{bmatrix}
$
\Rightarrow
$
\begin{bmatrix}
0.25 x_{2}\\
x_{1}+0.75 x_{2}
\end{bmatrix}
$
\Rightarrow
$
\begin{bmatrix}
25 \\
175
\end{bmatrix}
$
\boxed{\begin{align*}
0.25 x_{2} &= x_{1}\\
x_{1} + 0.75 x_{2} &= x_{2}
\end{align*}}
$
 A: The steady state is the population split that is an eigenvector of the transition matrix with eigenvalue 1, that is:
$$\left(\begin{matrix} 0 & 1/4 \\ 1 & 3/4 \end{matrix}\right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \left( \begin{matrix} x \\ y \end{matrix} \right)$$
This leads you to the equation
$$y = 4x$$
and you also know that
$$x + y = 200$$
and solving these jointly gives $x=40$, $y=160$.
A: I assume that what you did with the calculator is to keep multiplying by the matrix until things appeared to stabilize.  
Very streamlined versions of what you did are often used in practice.  So the idea was a good one, and shows that you understand the meaning of long-term distribution. The idea is not wrong, but it is almost certainly not what this part of the problem expects you to do.
In a first Linear Algebra course, with a small ($2 \times 2$) matrix, you would be expected to calculate an eigenvector for eigenvalue $1$, as described by Chris Taylor.  
Possibly your approach would get some part marks.  But you have been taught about eigenvalues and eigenvectors. This part of the question is probably designed to see whether you are aware of the connection between eigenvectors for eigenvalue $1$ and the long-term distribution, and can do the requisite eigenvector calculation.
A: When working with probabilities one condition that is often overlooked is that all probablilities add up to 1. You are not done with the steady state when you get to the boxed equations, you need to use the sum of probabilities are 100%, this is sneaked into this problem as the "extra" condition $x+y = 200$, but it is important to think about ( for other problems ) that the probability interpretation is that all probability adds up to 1.
