Actually, I would just ignore the "matrix" and look at the columns and rows as vectors (that is, after all, what the problem is asking about).

The rows are <-2, 1, -3, 6, 1>, <1, 1, 1, -2, 1>, and <2, 3, 1, -2, 4>. If those are "independent" then they form a basis themselves. To see if they are independent, use the definition:

look at the equation a<-2, 1, -3, 6, 1>+ b<1, 1, 1, -2, 1>+ c<2, 3, 1, -2, 4>= <0, 0, 0, 0, 0>. If the only solution is a= b= c= 0 (very likely, since you have an "overdetermined" system- more equations than unknowns) they are a basis. If not, you can solve for one equation as a combination of the other two.

Similarly for the columns. They are <-2, 1, 2>, <-1, 1, 3>, <-3, 1, 1>, <6, -2, -2>, and <1, 1, 4>. Again, look at the equation a<-2, 1, 2>+ b<-1, 1, 3>+ c<-3, 1, 1>+ d<6, -2, -2>+ e<1, 1, 4>= <0, 0, 0>. Now you have three equations in a, b, c, d, and e. You will be able to solve for some of them in terms of the others- specifically, you should be able to solve for three of the unknowns in terms of the other 2. And that means that you can replace those two vectors with a linear combination of the three.

This is really the same as "row reducing" the matrix but I think it is best to stay as close to the fundamental definitions.