Prove that $B=\{(1,1,4), (1,1,3), (1,-1,2)\}$ is a basis of vector space $\Bbb R^{3}$. Prove that $B=\{(1,1,4), (1,1,3), (1,-1,2)\}$ is a basis of vector space $\Bbb R^{3}$. 
To prove this, I have to prove whether this set is linearly independent and whether the vectors in $B$ span $\Bbb R^{3}$. 
$1)$ $\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=0$ 
$(\alpha, \alpha, 4\alpha)+(\beta,\beta,3\beta)+(\gamma,-\gamma,2\gamma)=0$
The system 
$\alpha+\beta+\gamma=0$
$\alpha+\beta-\gamma=0$ 
$4\alpha+3\beta+2\gamma=0$ has the solution $\alpha=\beta=\gamma=0$
So these vectors $(1,1,4), (1,1,3), (1,-1,2)$ are linearly independent.
$2)$ Now I'm not sure how to check these vectors span $\Bbb R^{3}$. I was told to do this:
$\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=(x,y,z)$ $\Rightarrow$ 
$\alpha+\beta+\gamma=x$ 
$\alpha+\beta-\gamma=y$ 
$4\alpha+3\beta+2\gamma=z$ 
I'm showing that any vector in $\Bbb R^{3}$ can be represented as linear combination of those three given vectors, is that right? 
What should I do with this system of equations?
 A: 
$2)$ Now I'm not sure how to check these vectors span $\Bbb R^{3}$. I was told to do this:
$\alpha(1,1,4)+\beta(1,1,3)+\gamma(1,-1,2)=(x,y,z)$ $\Rightarrow$ 
$\alpha+\beta+\gamma=x$ 
$\alpha+\beta-\gamma=y$ 
$4\alpha+3\beta+2\gamma=z$ 
I'm showing that any vector in $\Bbb R^{3}$ can be represented as linear combination of those three given vectors, is that right? 
What should I do with this system of equations?

Now you have to solve this system for $\alpha,\beta,\gamma$ (see below). If this system has a solution, for arbitrary $x,y,z$; then you have shown that you can find suitable values of $\alpha,\beta,\gamma$ to 'make' any $(x,y,z)$ as a linear combination of the given vectors. They then span $\mathbb{R^3}$.
Note that this is the hard way, staying close to the definition of 'basis', checking if the vectors:


*

*are linearly independent;

*span the space.


When or if you have seen some relevant properties/theorems, you'll know that 3 linearly independent vectors will always span a 3-dimensional vector space - a lot easier.

For future reference and perhaps you can check:
$$
\left\{ 
\begin{array}{l}
\alpha+\beta+\gamma=x \\ 
\alpha+\beta-\gamma=y \\ 
4\alpha+3\beta+2\gamma=z
\end{array}
\right. \iff
\left\{ 
\begin{array}{l}
\alpha=-\tfrac{5}{2}x-\tfrac{1}{2}y+z \\ 
\beta=3x+y-z \\ 
\gamma=\tfrac{1}{2}x-\tfrac{1}{2}y
\end{array}
\right. 
$$
A: Three linearly independents vectors in $\mathbb R^3$ obviously span $\mathbb R^3$.
You just need to verify the linear independence.
