Calculate total number of matrices of all orders which contain $2013$ elements
My Try:: By Simple Guessing wecan say that there are two matrices of order
$(1\times 2013)$ and $(2013 \times 1)$
But my question is: How can I check other matrices exist or not, and How can I check whether numbers like $2013$ and $2011$ are prime or not.
You will need to factor $2013$. A simple start is that because the sum of the digits of $2013$ is divisible by $3$, the number $2013$ is divisible by $3$.
Divide. We get $671$. Is this further divisible? You can test various primes. There is no point in testing $5$; $7$ doesn't work; and we get success at $11$. Divide. We get $61$.
So $2013=3\cdot 11\cdot 61$. Now can you find the required number? As a check on your further work, it will turn out that there is a total (including the ones you found) of $8$ possibilities.
Remark: The answer to your problem is, as was probably clear to you, the number of positive divisors of $2013$. It can be shown that if $n$ has prime power factorization $$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},$$ where the $p_i$ are distinct primes, then $n$ has $$(a_1+1)(a_2+1)\cdots (a_k+1)$$ distinct positive divisors.
You do not need this result to solve the $n=2013$ problem, a simple list will do the job.