An n by n square matrix A is per definition singular if it is not invertible. There are several ways of determining this. As Adrian Keister pointed out, A is singular if and only if it's determinant is equal to zero, which can relatively easy be computed for n=1,2,3 in the general case(by hand). Using this method for lager n generally becomes rather tedious.
Now, if you are familiar with the "standard" way of calculating the inverse of a matrix A (row reducing [A| i] to [i|A^-1] ), this process requires that A is row equivalent to the n by n identity matrix i. So A is singular if and only if it is not row equivalent to the n by n identity matrix i. This test is easy by hand for n=1,2,3,4, 5 in the general case, and might be tedios for larger n.
Furthermore, if you have learned about how to determine if a set of column vectors are linearly independent, you know that they are linearly independent if and only if the matrix consisting of each vector as a column is row equivalent to the identity matrix, that is; their span is equal to n if and only if this mentioned matrix is row equivalent to i. If we think about this matrix as a complete unit, the dimension of it's colunmn space (it's rank) is equal to n if and only if the matrix is row equivalent to i. Combining these we get that a n by n square matrix A is singular (non-invertible ) if and only if its rank is not equal to n, the amount of columns of A. And since it's rank can't be greater than n it thus has to be less than n.
If the given matrix A has a row or column consisting entirely of zeros it is automatically singular, in correspondence with the previous points mentioned.
There is a huge bunch of things that are equivalent to a n by n matrix A being invertible, see for example http://mathworld.wolfram.com/InvertibleMatrixTheorem.html
If either of these conditions fails to hold for a square matrix A, then it is not invertible.