Answer
See answers below
Work Step by Step
a) Assume that $A=\begin{bmatrix}
1 & 1 & -1 & 2 \\
2 & 1 & 3 & -4\\
1 & 2 & -6 & 10
\end{bmatrix} $
then we reduce $A$: $\begin{bmatrix}
1 & 1 & -1 & 2 \\
2 & 1 & 3 & -4\\
1 & 2 & -6 & 10
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & -1 & 2 \\
0 & -1 & 5 & -8\\
0 & 1 & -5 & 8
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & -1 & 2 \\
0 & -1 & 5 & -8\\
0 & 0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 1 & -1 & 2 \\
0 & 1 & -5 & 8\\
0 & 0 & 0 & 0
\end{bmatrix}$
The basic for rowspace $A$ is $\{(1,1,-1,2);(0,1,-5,8)\}$
Hence $S$ is basic of given subspace.
b) Assume that column of $A$ formed by given vectors,
then $\begin{bmatrix}
1 & 2 & 1 \\
1 & 1 & 2\\
-1 & 3 & -6\\
2 & -4 & 10
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 1 \\
0 & -1 & 1\\
0 & 5 & -5\\
0 & -8 & 8
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 1 \\
0 & -1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 1 \\
0 & 1 & -1\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$
We notice that the first and second bolumns are independent. The basic for colspace $A$ is: $\{(1,1,-1,2);(2,1,3,-4)\}$
Hence $S$ is a basic of given subspace.