determine whether $v_1=(1,1,2)$ , $v_2=(1,0,1)$ and $v_3=(2,2,3)$ span $\mathbb{R}^3$ determine whether $v_1=(1,1,2)$ , $v_2=(1,0,1)$ and $v_3=(2,2,3)$ span $\mathbb{R}^3$
I think that it must prove that it linear Independence 
let   $α_1v_1+α_2v_2+α_3v_3=(0,0,0)$
so what next , is this a correct way or I'm completely wrong 
please guide me to understand this
thanks
 A: There is a theorem in linear algebra which says that any set of $n$ linearly independent vectors in an $n$-dimensional space is a basis for that space.  $\Bbb R^3$ is $3$-dimensional (I assume you've proven this at some point this semester?) and you've got $3$ vectors.  Thus those vectors form a basis for $\Bbb R^3$ if and only if they are linearly independent.  Below I've provided a few methods of determining whether or not they are linearly independent.
Method #1
Construct the following matrix: $$\begin{bmatrix} 1 & 1 & 2 \\ 1 & 0 & 2 \\ 2 & 1 & 3 \end{bmatrix}$$
Now use Gauss-Jordan elimination to determine the rank of the matrix.  If it is $3$ (the number of columns), then the columns are linearly independent.
Method #2
Construct the same matrix as above.  Find the determinant of that matrix.  If it is nonzero, then the columns are linearly independent.
Method #3
Look at the equation $a_1(1,1,2) + a_2(1,0,1) + a_3(2,2,3) = (0,0,0)$.
This is equivalent to the set of scalar equations: $$\begin{cases} a_1+a_2+2a_3 = 0 \\ a_1+2a_3=0 \\ 2a_1 + a_2 +3a_3 = 0\end{cases}$$
We can immediately see that $a_1 = -2a_3$.  Plugging this into the first equation we get: $(-2a_2)+a_2 +2a_3 = -a_2+2a_3 = 0 \implies a_2 = 2a_3$.  Plugging both of these into the third equation we get $2(-2a_3)+(2a_3)+3a_3=0 \implies a_3=0$.  Plugging this back into $a_1=-2a_3$ and $a_2=2a_3$, we see that $a_1=a_2=a_3=0$.  Therefore these $3$ vectors are linearly independent.
Method 4
We know that $(1,0,0), (0,1,0), (0,0,1)$ is a basis for this space.  So if each of these standard basis vectors can be represented by a linear combination of your three vectors $v_1, v_2, v_3$, then $\operatorname{span}((1,0,0), (0,1,0), (0,0,1)) \subseteq \operatorname{span}(v_1, v_2, v_3)$.  Because there are $3$ of them, we can see that $\{v_1, v_2, v_3\}$ will thus be a basis for $\Bbb R^3$ and thus linearly independent.
Method 5
Take the exterior product of the three vectors, $v_1 \wedge v_2 \wedge v_3$.  If it is nonzero, then the three vectors are linearly independent.
A: You have three vectors and $dim(\mathbb{R}^{3}) = 3$. So all you have to test for is linear independence. You can do this by constructing a matrix $A$ as JohnD suggested. Then just take $det(A)$. The vectors form a basis iff $det(A) \neq 0$.
A: Put their columns in a matrix $A$. You want to know if for all vectors $\mathbf{b}=(a,b,c)$ if $A\mathbf{x}=\mathbf{b}$ has a solution. Do you know how to answer that question?
A: You are on the right way. That yields
$$\alpha_1 (1,1,2) + \alpha_2 (1,0,1) + \alpha_3 (2,2,3) = (0,0,0)$$
$$
\left( \begin{array}{ccc}
1 & 1 & 2 \\
1 & 0 & 2 \\
2 & 1 & 3 \end{array} \right)
\left( \begin{array}{c}
\alpha_1 \\
\alpha_2 \\
\alpha_3 \end{array} \right)
=
\left( \begin{array}{c}
0\\
0\\
0\end{array} \right)$$
which is a matrix equation. You can solve the system of linear equation or just see whether the determinant is $0$. Since this system must have a trivial solution, either it has a unique solution (hence the trivial one and thus the set is linearly independent) or infinitely many solutions (hence the set is linearly dependent).
