you have 4 vectors:
v1=(2,1,-1), v2=(-m,-1,3), v3=(-3,2,m+1), v4=(1,2,1)
a) for which values of m do the vectors span R^3
b) the vector w=(m+1, m-1, 1). for which m is there a solution to:
x1v1 + x2v2 + x3v2 + x4v4 = w ?
the definition of spanning - every vector in R^3 can be made of those 4 vectors.
The Attempt at a Solution
a) i made the following matrix where the first few coloums are v4,v1,v3,v4 and the last one is some vector (a,b,c):
[1 2 -3 -m a] [2 1 2 -1 b] [1 -1 (m+1) 3 c] and reduced it to: [1 2 -3 -m a] [0 -3 8 (2m-1) b-2a] [0 0 (m-4) (4-m) c-b+a]
[0 0 0 0 alpha] where alpha is any scalar. since the only row that can possibly look like this is the third one where m=4 then the answer is that they span R^3 for every m except 4.
Is that right?
b) here's where i got confused, if m doesn't equal 4 then there's a solution because according to a) the vectors v1,v2,v3 and v4 span R^3 which includes w. and if m=4 then there's no solution because for w:
c-b+a = 3 (not 0)
so the answer for b should be the same as a). is that true? it doesn't seem likely that there isn't some trick or something.