For part a, my work is:
Reduced
The system is inconsistent and has no solution so this set does not span R^3?
Part b: same work resulting in the following reduced matrix
So this would also not span R^3 since there are no solutions?
Part c:
So this set does span R^3?
Does anyone know if this is the way this problem should be done? Thank you for your help.
Since two vectors can't span a three dimensional space, and you said that you understood that, I don't know why you did that.
But that method can be used on the others: {(2,2,3),(-1,-2,1),(0,1,0)} spans if and only if there exist x, y, z such that x(2,2,3)+ y(-1,-2, 1)+ z(0, 1, 0)= (a, b, c) for any three numbers a, b, c. We can write that as 2x- y= a, 2x-2y+ z= b, 3x+ y= c. Adding the first and third equations, 5x= a+c or x= (a+c)/5. Then the third equation gives y= c- 3x= c- (3/5)a- (3/5)b. Put that into the second equation to see if you can solve it for z.
Of course that is the same as reducing the augmented matrix
Actually, a, b, and c are irrelevant here. such x, y, z will exist for all a, b, c if and only if row reducing the matrix
does not give a row consisting entirely of 0s. With four vectors you get a four rows and you only need three of them not consisting of 0s.
In my text, that is how the work is shown in the examples. When there was a set of 3 vectors for R^4, work was still shown in that format. So that's why I was so confused. But otherwise, my work is sufficient? Simply reduce and make sure there are enough nonzero rows. (ex. for a set of vectors to span R^3, there must be three nonzero rows after the matrix is reduced?)
For part d, the matrix created was already reduced and all rows were nonzero (3 rows x 4 columns), so that would also span a 3-D space? Therefore, a and b are the only sets that would not. I just want to show enough work for my answers. Thank you!