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