Is the vector $(3,-1,0,-1)$ in the subspace of $\Bbb R^4$ spanned by the vectors $(2,-1,3,2)$, $(-1,1,1,-3)$, $(1,1,9,-5)$? Is the vector $(3,-1,0,-1)$ in the subspace of $\Bbb R^4$ spanned by the vectors $(2,-1,3,2)$, $(-1,1,1,-3)$, $(1,1,9,-5)$?
 A: To find wether $(3,-1,0,-1)$ is in the span of the other vectors, solve the system:
$$(3,-1,0,-1)=\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5)$$
If you get a solution, then the vector is the span. If you don't get a solution, then it isn't.
It's worth noting that the span of $(2,-1,3,2), (-1,1,1,-3), (1,1,9,-5)$ is exactly the set $\left\{\lambda_1(2,-1,3,2)+\lambda _2(-1,1,1,-3)+\lambda _3(1,1,9,-5):\lambda _1, \lambda _2, \lambda _3\in \Bbb R\right\}$

Alternatively you can consider the matrix $\begin{bmatrix}2& -1 &3 & 2\\ -1 &1 &1 &-3\\ 1 & 1 & 9 & -5 \\ 3 &-1 &0 & -1\end{bmatrix}$. Compute its determinant. If it's not $0$, then the four vectors are linearly independent. If it is $0$ they are linearly dependent. What does that tell you?
A: You have this coefficient(augmented) matrix: $\begin{pmatrix}
2&-1&1&3\\
-1&1&1&-1\\
3&1&9&0\\
2&-3&-5&-1
\end{pmatrix}$
Reduce it to the row echelon form and check whether it is consistent.
You basically need to check whether after row reduction its last non-zero row has a pivot in the last column or not.If it has in the last column then the equation set is inconsistent so you dont have a solution.Else you have solution(infinitely many) implying that the vector belongs to the span of the given three vectors.
A: if (3,-1,0,-1) be in span {(2,-1,3,2),(-1,1,1,-3),(1,1,9,-5)} then we must have
 (3,-1,0,-1)=$c_{1}$ (2,-1,3,2)+ $c_2$ (-1,1,1,-3)+ $c_3$ (1,1,9,-5)
so this system must have solution:
$2c_1 -c_2+c_3=3$
$-c_1 +c_2+c_3=-1$
$3c_1 +c_2+9c_3=0$
$2c_1 -3c_2-5c_3=-1$ 
since 3 first relation has unic answer and 3 end realations too and they cant have equal answer (then this system has not solution and (3-10-1) 
