Constructing a matrix of order $3\times 3$ such that the limiting matrix also exists in which all the rows are not the same. Let $$ A = \left[\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} \right] $$ be a matrix where $a_{ij}\in{[0,1]}$, $\sum_{j=1}^{3}a_{ij}=1$ (for $i=1,2,3$), and $a_{ij}\in{(0,1)}$ for $i\neq{j}$. Then how do we construct a matrix $A$ such that $A_{\infty}:=lim_{n\rightarrow{\infty}}A^n$ exists and $A_{\infty}$ does not have all the rows equal?
Or, how do we construct a square matrix of order bigger than 2 (of non-negative numbers ) having at least two elements in any row are not zero and the sum of each row is 1?
I tried 
$$ 
A = \left[\begin{matrix} 0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0 \end{matrix} \right] 
$$
And, for this matrix $A_{\infty}$ does not exist. 
If we try 
$$ 
A = \left[\begin{matrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{matrix} \right] 
$$
Then, $A_{\infty}$ exists but the problem is $A_{\infty}$  has all the rows equal. So, I could not come up with a matrix satisfying both properties.
 A: For your first matrix,
$$
A = \begin{bmatrix} 0 & 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{bmatrix},
$$
$A_\infty$ does exist and is equal to 
$$
A = \begin{bmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3\end{bmatrix}.
$$
A matrix with the properties you want does not exist, at least in the first statement of the question. This follows from results related to Markov chains, as the matrices you describe are transition matrices for Markov chains. Specifically, an $n\times n$ matrix $A$ such that $a_{ij} \in[0,1]$ and $\sum_{j=1}^n a_{ij}=1$ for each $n$ describes the transition of a Markov chain with $n$ states, where $a_{ij}$ is the probability of moving from state $i$ to state $j$ in a time step. A Markov chain is called regular if every entry of $A^k$ is nonzero for some positive integer $k$ (intuitively that it is possible to move between any two states in some number of time steps). Your requirement that $a_{ij}\in(0,1)$ for $i\neq j$ implies that the chain is regular, as it is possible to pass between any two states in one time step if they are different states ($i\neq j$, so $a_{ij}\neq 0$) or two if they are the same state (as you can go from the $i$th state to any other state and then back to the $i$th state).
For every regular Markov chain, it is a theorem that there is a steady state vector i.e. a distribution among the states that remains constant (there are infinitely many, but we get a unique one by insisting that it is a probability vector, i.e. the sum of the entries is $1$). Moreover, each row of $A_\infty = \lim_{k\to\infty} A^k$ is equal to this steady state vector, and hence all the rows of $A_\infty$ coincide.
For the weaker requirement,

Or, how do we construct a square matrix of order bigger than 2 (of non-negative numbers ) having at least two elements in any row are not zero and the sum of each row is 1?

You can do this by finding a transition matrix for a Markov chain that is not regular. A simple example comes from a disconnected Markov chain (really, two independent Markov chains treated as one). E.g.
$$
A = \begin{bmatrix} 1/2 & 1/2 & 0 & 0 \\ 1/2 & 1/2 & 0 & 0 \\ 0 & 0 & 1/2 & 1/2 \\ 0 & 0 & 1/2 & 1/2 \end{bmatrix}.
$$
You can compute that $A^2 = A$ so that $A_\infty = A$.
