The rank of a matrix is equal to the span of the column space of the matrix or the number of linearly independent columns of the matrix.
It can also be said that the rank of a matrix is equal to the number of non-zero rows in the reduced row echelon form of a matrix.
Debbie's description is absolutly right and I will show you how to calculate the reduced row echelon form of a matrix.

To transform a matrix into the RREF it is only allowed to use so called elemental operations (I hope this is the correct English term):
1. Multiply every element of a row or column with a scalar:

|1 4 5|        |2 8 10|  (*2)
|3 3 1|    ~   |3 3 1 |
|5 3 9|        |5 3 9 |

2. Add the multiple of a row(column) to another row(column):

I:   |1 4 5|        |1  4   5| 
II:  |3 3 1|    ~   |0 -9 -14|  (II-3*I)
III: |5 3 9|        |5  3   9|

3. Swap rows or columns

|1 4 5|        |3 3 3|
|3 3 1|    ~   |1 4 5|
|5 3 9|        |5 3 9|

That's everything that is allowed to do, because these operations do not change the rank of the matrix.
Now I will describe the algorithm itself, afterwards I will give a short example:
The matrix is called A and its elements are a(i,j) where i = 1..m and j=1..n. The first thing to do is to make a(1,1)=1. If a(1,1)!=0 then it is very easy: Just divide the first row or column by its value. If a(1,1)=0 then you have to swap columns and(or) rows to get a non-zero value there.
The next thing to do is to make the rest of the first row and column to zero. We start with the rows: Every row k=2..m will be multiplied with a(k,1) times the first row.
The same has to be done with the columns (though here it is every column l=2..n must be multiplied with a(1,l) times the first column.
The above algorithm has to be repeated for the (m-1)x(n-1) matrix, leaving the first row and column alone. It has to be repeated till the RREF is achived.
The number of non-zero rows or columns is the rank of the matrix.

Example:

I:   |1 2 3|   |1  2  3|             |1  0  0|   |1  0  0| 
II:  |2 3 4| ~ |0 -1 -2| (II-2*I)  ~ |0 -1 -2| ~ |0  1  2|
III: |3 4 5|   |0 -2 -4| (III-3*1)   |0 -2 -4| ~ |0 -2 -4|

I:   |1  0  0|            |1 0  0|   |1 0  0|   |1 0 0|
II:  |0  1  2| (*(-1)) ~  |0 1  2| ~ |0 1  0| ~ |0 1 0|
III: |0 -2 -4|            |0 0 -4| ~ |0 0 -4| ~ |0 0 1|

The last step is not needed as one can already see from the previous matrix that the rank R(A)=3. If you want to check if you understood the algorithm, just use the following example:
|1  2 0 0|
|1 -1 1 0|
|1  0 2 2|

The rank of this matrix is 3. Calculate it yourself!

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