Ways to find basis of a span of matrices 
let $W= $$\left(\begin{matrix}
b & -a & 3a+b \\
c & 0 & a+2c \\
-3c & c & b \\
\end{matrix} \right)
$ $ \in M_3( \Bbb R)$ and $a,b,c \in \Bbb R$  I would like to know ways to find the basis of W using basic methods so I can have more options while practicing.

for this case I first found the span $W = a$ $\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
$
$+b  \left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
$
$+c \left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$
so from here the span is $W=$Sp$\{$$\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
$ ,
$\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
$ ,
$\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$ $\}$
and a basis needs to span and be linear independent so we will check linear independence by using coordinates vector by using the standard basis for $ M_3( \Bbb R)$ ($dim M_3( \Bbb R)=9$)
then using matrix elementary row operations for the following matrix
$\left(\begin{matrix}
0 & -1 & 3 & 0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &1 \\
0 & 0 & 0 & 1 & 0 & 2 & -3 & 1 & 0 \\
\end{matrix} \right)
$ $R_1 \iff R_2$
$\left(\begin{matrix}
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &1 \\
0 & -1 & 3 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 2 & -3 & 1 & 0 \\
\end{matrix} \right)
$
we have 3 opening elements and no zero rows so it is linearly independent and we found the span so it is a basis for W.
this is the way I know and I am just wondering if there are any other methods (that are suitable for linear algebra 1 course)
Thank you!

From the matrices
$$\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
 ,\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
 ,
\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$$
you ought to choose a linear combination
$$K\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
+L\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
+M\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$$
and equate to the zero matrix
$\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \right)
$ to deduce that the coefficients all $K,L,M$ are zero, which will imply that they are linearly independent.
This leads you to use
$$\left(\begin{matrix}
L & -K & 3K+L \\
M & 0 & K+2M \\
-3M & M & L \\
\end{matrix} \right)=
\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \right),
$$
with that goal in mind.
