Rank of a matrix. Let a non-zero column matrix $A_{m\times 1}$ be multiplied with a non-zero row matrix $B_{1\times n}$ to get a matrix $X_{m\times n}$ . Then how to find rank of $X$?
 A: Hint: To find the rank of a matrix, one usually row (or column) reduces it. In your case, all the rows (or columns) are multiples of one row (column). What can you conclude about the reduced matrix?
A: Hint:  You always have
$$rk(AB)\leq min\{rk(A),rk(B)\}$$
It would be helpful to convince yourself of this fact. @DennisGulko's answer gives you an idea how.
Now can the rank of $AB$ be zero? That would mean that the matrix has only zero-entries...
Edit: I assume the entries of the matrix are elements of a field, i.e there are no zero-divisors.
A: Let me discuss a shortcut for finding the rank of a matrix . 
Rank of a matrix is always equal to the number of independent equations .
The number of equations are equal to the number of rows and the variables in one equation are equal to number of columns  .
Suppose there is a 3X3 matrix with elements as :
row 1 : 1 2 3 
row 2 : 3 4 2 
row 3 : 4 5 6 
So there will be three equations as
x + 2y + 3z = 0    -1 
3x + 4y + 2z = 0   -2 
4x + 5y + 2z = 0   -3 
Any of the above equation cannot be obtained by adding or subtracting two equations or multiplying or dividing a single equation by a constant . So there are three independent equations . So rank of above matrix is 3 .
Consider another matrix of order 3X3 with elements as :
row 1 : 10 11 12
row 2 : 1  2  7
row 3 : 11 13 19
equations :
 10x + 11y + 12z =0   - 4
 x + 2y + 7z =0       - 5
11x + 13z + 19z =0    - 6 
equation 6 can be obtained by adding equations 4 and 5 . So there are only two independent equations . So rank of this matrix is 2 . 
This method can be applied to matrix of any order .
