Finding a basis (linear algebra)

Let T:R^3->R^3 have the indicated properties. Find a basis for the range of T. (2 part question)
a. Te1=(2,0,-1), Te2=(1,3,0), Te3=(-1,9,2)
b. T(1,1,0)=(2,0,1), T(0,2,-1)=(0,2,1), T(1,0,1)=(-4,6,1)

Notice that, in both cases, T is defined by the images of basis points. Since bases span the entire space, their images will span the entire range of T (this is straightforward to prove, if you want to give it a try). So, the sets, respectively for parts a and b:

{(2, 0, -1), (1, 3, 0), (-1, 9, 2)}
{(2, 0, 1), (0, 2, -1), (-4, 6, 1)}

form spanning sets for the ranges of the respective transformations. As we know, all spanning sets contain bases, so we just need to reduce these sets to linearly independent sets of the same span. There is a simple way to do so. We put the vectors in the given set into columns of a matrix, and row-reduce. Any column without a leading 1 in the row-echelon form can be discarded, leaving us with a linearly independent set of the same span (i.e. a basis for the range of T). Let's try it with the first one. We get the matrix:

(2 1 -1)
(0 3 9)
(-1 0 2)

We need to row-reduce this matrix. So:

(-1 0 2) ... (swap R1 and R3)
(0 3 9)
(2 1 -1)

(-1 0 2) ... (add 2 * R1 to R3)
(0 3 9)
(0 1 3)

(1 0 -2) ... (multiply R2 by -1)
(0 3 9)
(0 1 3)

(1 0 -2) ... (divide R2 by 3)
(0 1 3)
(0 1 3)

(1 0 -2) ... (subtract R2 from R3)
(0 1 3)
(0 0 0)

We are in row-echelon form, and the leading 1s occur in the first and second column. Thus, the third column vector (i.e. -1, 9, 2) is not necessary, meaning that {(2, 0, -1), (1, 3, 0)} is a basis for the range of T.

Try the second one yourself. You should arrive at the conclusion that one of the vectors in the set is also unnecessary (it doesn't matter which one, so long as one of them is dropped).