Find a basis for the subspace of $\mathbf{R}^4 $ spanned by the given vectors
(1,1,-3,-2), (3,0,3,-3), (3,-1,-2,13)
A problem that was similar involved reducing an augmented Matrix to RREF but that doesn't seem to work for this problem, possibly due to dimensional limitations since I tried the answer (1,0,0,0) , (0,1,0,0) and (0,0,1,0) and that didn't work.
The following possible answers are:
A:(1,1,-3,-2)
B:(1,1,-3,-2), (0,1,-4,-1) and (0,0,1,-5/3)
C:(1,1,-3,-2), (0,1,0,0)
You've got to search solutions for : $t_1(1,1,-3,-2)+t_2(3,0,3,-3)+t_3(3,-1,2,13)=(0,0,0,0)$
if $t_1=t_2=t_3=0$ is the only solution. your three vectors are a basis for the subspace. Otherwise, you have to look for numbers of free parameters of the set of solutions and to erase the vector(s) that are associated to those parameters. In your case :
$\begin{pmatrix} 1 & 3 & 3 & \\ 1&0&-1 \\-3&3&-2& \\ -2&-3&13 \end{pmatrix} \to$ gaussian elimination $\to \begin{pmatrix} 1 & 3 & 3 \\ 0 &-3 & -4 \\ 0 & 0 & -9 \\ 0 & 0 & 0 \end{pmatrix}$
Thus the vectors are independant and are a basis for the subspace.
Another answer is : you put the vectors in the rows of a matrix : $\begin{pmatrix} 1 & 1 & -3 & -2 \\ 3& 0 &3 &-3 \\3&-1&-2&13 \end{pmatrix}$ and you're proceeding to gaussian elimination. The result is : $\begin{pmatrix} 1 & 1 & -3 & -2 \\ 0& 1 & -4 & -1 \\ 0&0&-3&5 \end{pmatrix}$. The three rows of the last matrix are vectors that span the same subspace like your three starting vectors. you see immediatly that this is an independant set of vector, so a basis for your subspace.