Answer
See answers below
Work Step by Step
a) Assume that $A=\begin{bmatrix}
1 & -1 & 2 \\
5 & -4 & 1\\
7 & -5 & 4
\end{bmatrix} $
then we reduce $A$: $\begin{bmatrix}
1 & -1 & 2 \\
5 & -4 & 1\\
7 & -5 & 4
\end{bmatrix} \approx \begin{bmatrix}
1 & -1 & 2 \\
0 & 1 & -9\\
0 & 2 & -18
\end{bmatrix} \approx \begin{bmatrix}
1 & -1 & 2 \\
0 & 1 & -9\\
0 & 0 & 0
\end{bmatrix}$
The basic for rowspace $A$ is $\{(1,-1,2);(0,1,-9)\}$
Hence $S$ is basic of given subspace.
b) Assume that column of $A$ formed by given vectors,
then $\begin{bmatrix}
1 & 5 & 7 \\
-1 & -4 & -5\\
2& 1 & -4
\end{bmatrix} \approx \begin{bmatrix}
1 & 5 & 7 \\
0 & 1& 2\\
0& -9 & -18
\end{bmatrix} \approx \begin{bmatrix}
1 & 5 & 7 \\
0 & 1& 2\\
0& 0 & 0
\end{bmatrix}$
We notice that the first and second bolumns are independent. Since $m=3$, the basic for colspace $A$ is: $\{(1,-1,2);(5,-4,1)\}$
Hence $S$ is a basic of given subspace.