Basis vector find Let $\alpha_1=[ 2,1,3,0] $
$\alpha_2=[ 1,1,1,-1] $, $\alpha_3=[ 2,-1,5,4] $, $\alpha_4=[ 1,2,0,-3] $, $\alpha_5=[ 3,1,6,1] $
be vectors from $\mathbb{R}^4$ . From vectors system ($\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5 $) choose basis of vector space $V=lin(\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5)\subset\mathbb{R}^4$ spanned by those vectors.
I need help with creating a proper matrix for this question.
 A: Consider the matrix
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
and perform Gaussian elimination on it:
\begin{align}
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1/2 & −2 & 3/2 & -1/2\\
0 & -1/2 & 2 & -3/2 & 3/2\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
The columns where a pivot is found are linearly independent and the others can be written as linear combination of them. So a basis is
$$
\{\alpha_1,\alpha_2,\alpha_5\}
$$
It's not the only solution, of course, but you can see that $\alpha_5$ will be in any basis extracted from that set, because it is not a linear combination of the other four vectors. If you go on with backwards elimination,
\begin{align}
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 3 & -1 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
you see that
\begin{align}
\alpha_3&=3\alpha_1-4\alpha_2\\
\alpha_4&=-\alpha_1+3\alpha_2
\end{align}
because elementary row operation don't change linear relations between columns.
A: Since
$$\mathrm{rank}\begin{pmatrix}2 & 1 & 2 & 1 & 3\cr 1 & 1 & −1 & 2 & 1\cr 3 & 1 & 5 & 0 & 6\cr 0 & −1 & 4 & −3 & 1\end{pmatrix}=3$$
the vector subspace $\mathrm{span}\{\alpha_1,\cdots,\alpha_5\}$ has dimension $3.$ So, you need to find three linearly independent vectors.
Now, since 
$$\mathrm{det}\begin{pmatrix}2 & 1 & 3\cr −1 & 2 & 1\cr 5 & 0 & 6\end{pmatrix}=24+5-30+6=5\ne 0$$ we have that $\{\alpha_3,\alpha_4,\alpha_5\}$ is a linearly independent set of vectors. So we have got a basis.
