Linear Algebra: Prove there exists a state matrix The problem:

The attempt at solution:

Am I supposed to just choose random numbers (like I did) just for the sake of proving that there is a solution, or is there a predetermined way of doing this that gives me a better answer (or a range of answers)?
PS: This is for a Introductory Linear algebra class. 
 A: First, show $P$ has $1$ as an eigenvalue. Then let $v$ be a corresponding eigenvector. Show that $(P-I)v=0$. Think about which entries of $P-I$ are positive, which negative, and what that says about the entries of $v$. 
A: I will assume that you do not know anything about matrices except for the basic definitions.  The approach I will take is totally unfeasible for matrices much larger than $3 \times 3$, and is already unpleasant for $3 \times 3$.
From the example that you supplied, the matrix is probably supposed to be column-stochastic.  That is, the entries are non-negative, and the entries in any column add up to $1$.
We use more or less what you tried, except that I will let the entry on top left be $a$, and the entry on bottom right be $b$.  Then $P$ is the matrix
$$
\begin{pmatrix} a & 1-b \\
1-a & b
\end{pmatrix}
$$
(We could have let the entry on top right be $b$, and then the entry below would be $1-b$.  The calculation would be very similar to the one I will make, but marginally less symmetrical.)
We are looking for a column vector $X$ with non-negative entries such that $PX=X$.  We also want the entries not to be both $0$, though the problem did not mention that explicitly. Let the entries of $X$ be $x$ and $y$, with $x$ the top entry and $y$ the bottom one.
Calculate $PX$, with the usual procedure.  The top entry of the product is equal to $ax+(1-b)y$, and the bottom entry is $(1-a)x+by$.
We want this vector to be the same as the vector $X$.  So the entries must match.  We get the equations
$$ax+(1-b)y=x \qquad\text{and}\qquad (1-a)x+by=y$$
With standard "algebra" manipulation, these equations can be rewritten as
$$-(1-a)x + (1-b)y=0 \qquad \text{and}\qquad (1-a)x -(1-b)y=0$$
Note that these two equations say the same thing.  The second equation turns out to be the first, with all entries multiplied by $-1$.
So we only need to worry about satisfying the equation
$$-(1-a)x +(1-b)y=0$$
If $1-a$ and $1-b$ are both $0$, "any" $x$, $y$ is a solution.  If not both $1-a$ and $1-b$ are $0$, there is an obvious solution $x=1-b$ and $y=1-a$, and $x$ and $y$ are not both $0$.   
Since $a$ and $b$ are between $0$ and $1$, it follows that $x$ and $y$ are non-negative.
By the way, it is easy to check that if $k$ is any positive number, and if $x$, $y$ is a solution, then $kx$, $ky$ is also a solution.  Thus we can, by choosing $k$ appropriately, make $x$, $y$ satisfy some further condition, such as $x+y=1$.
Finally, please remember that the approach I took becomes very unpleasant in larger cases.  I wrote out a detailed solution to show that one can solve the problem with very little Linear Algebra background.  But in general, you will need the machinery referred to in other solutions. 
