Lets say we have a 5x6 matrix with rank=2
This means that there will be 2 leading ones and 4 parameters.
Does this mean that there will be 2 basic solutions to this matrix, or will each parameters count as a basic solution.
I'm confused on how to find the number of basic solutions.
Is it rank = # of basic solutions, which in this case = 2
columns - rank = # of basic solutions, which in this case would = 4
If we are looking at a homogeneous system of linear equations, and putting the coefficients into a matrix without an augmented column of $0$s, then the number of free parameters will be the number of basic solutions to the system.
If $t_1, \ldots, t_k$ are the free parameters, then we can get a basis of solutions in the following way. First, let $t_1 = 1$ and $t_2 = t_3 = \ldots = t_k = 0$. This gives us the first basic solution. Then, let $t_2 = 1$ and $t_1 = t_3 = \ldots = t_k$. This is your second basic solution. Keep going, getting more basic solutions by making all but one free parameter $0$, and the one remaining free parameter $1$. These solutions will form a basis for the space of solutions to the system of equations. Note that there are $k$ of them.