<s>[INST] Linear dependence and independence solution. How was it found? I am reading this text:

I feel like my text skipped a step to find the coefficients/solutions. Did they write a system of linear equations and use Gaussian elimination? How did they do this?
 [/INST] In order to see whether these vectors are dependent examine the system:
 $a(1,3,1)+b(0,1,2)+c(1,0,-5)=(0,0,0)$
which is equivalent with
$$\begin{pmatrix}
1 & 0 & 1\\
3 & 1 & 0\\
1 & 2 & -5
\end{pmatrix}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$$
then you can use elimination to find $a,b,c$
$$\begin{pmatrix}
1 & 0 & 1\\
3 & 1 & 0\\
1 & 2 & -5
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
1 & 2 & -5
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 2 & -6
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 0 & 0
\end{pmatrix}$$
So the above system is equivalent to the following
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & -3\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}\Leftrightarrow \begin{cases} a+c=0 \\ b-3c=0 \end{cases}\Leftrightarrow \begin{cases} a=-c \\ b=3c \end{cases}\Leftrightarrow \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}=\begin{pmatrix}
-c\\
3c\\
c
\end{pmatrix}=c\begin{pmatrix}
-1\\
3\\
1
\end{pmatrix} $$
And we can choose $c=1$ so $(a,b,c)=(-1,3,1)$
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