Using rref to give number of solutions (which is either 0, 1 or oo)

The matrix eqn A X = B, gives augmented matrix (A | B) which is turned
into rref form.

0: If the last column has a pivot, the system is not consistent. Another
way of waying this is the rank of the augmented matrix > the rank of the
coefficient matrix

1: If it is consistent and the number of pivots = number of unknowns

oo: If it is consistent and the number of pivots < number of unknowns.

Basically on page 12 of the text.

Solutions in the unique case are easy:
[ 1 0 0 | 3 ]
[ 0 1 0 | 5 ]
[ 0 0 1 | 0 ] 
The coefficient part is the identy, so the answer is just
x1 = 3, x2 = 5, x3 = 0; we will write this as a column vector
[ 3 ]
[ 5 ]
[ 0 ]

Solutions to the oo case are broken into two pieces. The particular
solution which makes the variables with columns with no pivot zero
and the homogenuous solution which gives an infinite family of solutions
for each variable with a column with no pivot. For example
[ 1 2 0 3 | 3 ]
[ 0 0 1 2 | 6 ]
has particular solution (set x2 = x4 = 0) so x1 = 3 and x3 = 6 this
gives the particular solution
[ 3 ]
[ 0 ]
[ 6 ]
[ 0 ]
Next we look at the homogenuous problem
[ 1 2 0 3 | 0 ]
[ 0 0 1 2 | 0 ]
WE get one infinite family for each non-pivot column:
column 2: Make x2 = 1 and x4 = 0 so x1 = -2 and x3 = 0
   [ -2 ]
   [ 1  ]
C1*[ 0  ]
   [ 0  ]
column 4: Make x2 = 0 and x4 = 1 so x1 = -3 and x3 = -2
   [ -3 ]
   [ 0  ]
C2*[ -2  ]
   [ 1  ]
All solutions can be written
[ 3 ]      [ -2 ]      [ -3 ]
[ 0 ]      [ 1  ]      [ 0  ]
[ 6 ] + C1*[ 0  ] + C2*[ -2 ]
[ 0 ]      [ 0  ]      [ 1  ]