Answer
See answer below
Work Step by Step
a) We obtain $\begin{bmatrix}
-3\\
1\\
1
\end{bmatrix} \approx \begin{bmatrix}
-3\\
0\\
0
\end{bmatrix}\approx\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}$
We notice that there is only one row vector which is equivalent to basis of row space of $A$ with $n=1$. So, the basis for row-space is of $A$ equal to $\{(1)\}$.
b) We notice that just one column $A$is dependent $\{(1)\}$ . So, the basis for column-space of $A$ is equal to $\{(-3,1,7)\}$ with $m=3$.
