Determine whether the set S = {v1,v2,v3} = {(1,-2,2) , (3,-5,3) , (-1,1,1) } spans R^3.

If the set does not span R^3, give a geometrical description of the subspace that it does span.

Here is my work but I think its wrong:

Let a vector u = (u1,u2,u3) in R^3.

Are there scalars c1,c2,c3 such that U = c1u1,c2u2,c3u3 is that system with augmented matrix containing each vector as a column consistent?

Let A = $\displaystyle \begin{array}{ccc}1&3&-1\\-2&-5&1\\2&3&1\end{array}$

Let C = $\displaystyle \begin{array}{ccc}c1\\c2\\c3\end{array}$

Let U = $\displaystyle \begin{array}{ccc}u1\\u2\\u3\end{array}$

Is there a solution C in R^3 to Ac = u for every u in R3 ?

by finding the det A using cofactor expansion I found the determinate to be -24 which is not 0. Therefore Ac = u has a solution in c in R^3 for every u in R^3

That would mean this set spans the subspace?

My notes from class are :

The following three statements are equivalent:

1) A is invertible

2) Ac = u has a solution c in R^n for every u in R^n

3) det A is not equal to 0

My second question:

Determine whether or not the set S = {v1,v2,v3} = {(1,-2,1) , (2,-3,5) , (1,-3,-2)} is linearly dependent or independent. If the set is dependent, then express one of the vectors in the set as a linear combination of the other vectors in the set.

According to my notes the set S = { V1,V2,....Vk } is linearly dependent if the equation c1V1 + c2V2 + .... + ciVk = 0 vector has a solution where not all of ci's are equal to zero.

In laymans terms there needs to be scalars c1,c2,...ck such that when multiplied with v1,v2,...vk you get the zero vector AND at least one of these scalars cannot be 0. The set S can only be independent if the trivial solution is the only solution to

c1V1 + c2V2 + ... + ck-1Vk-1 + ckVk = zero vector

I took my three vector and the zero vector and made the augmented matrix containing v1,v2,v3 and the zero vector as its columns. I then put that system in reduced row echelon form and got the following matrix:

$\displaystyle \begin{array}{cccc}1&0&3&0\\0&1&-1&0\\0&0&0&0\end{array}$

by looking at the matrix you can see that:

c1 + 3(c3) = 0

c2 - c3 = 0

this is our system of linear equations

I then let c3 = t and using back substituion I found c2 = t and c1 = -3t

then letting t = 1 I found c3 = 1, c2 =1, c1 = -3

you can conclude that -3V1 + V2 + V3 = 0

V1 = 1/3(V2) + 1/3(V3)

V2 = 3V1 - V3

V3 = 3V1 - V2

I can conclude that S is linearly dependent.

How does my work look? Thanks in advance