Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
-1 & 6 & 2 & 0\\
3 & 3 &1 & 5 \\
7 & 21 & 7 & 15
\end{bmatrix}$
We obtain rowspace $A=\begin{bmatrix}
-1 & 6 & 2 & 0\\
3 & 3 &1 & 5 \\
7 & 21 & 7 & 15
\end{bmatrix} \approx \begin{bmatrix}
-1 & 6 & 2 & 0\\
0 & 21 & 7 & 5 \\
0 & 63 & 21 & 15
\end{bmatrix} \approx \begin{bmatrix}
-1 & 6 & 2 & 0\\
0 & 21 & 7 & 5 \\
0 & 0 & 0 & 0
\end{bmatrix} \\
rowspace (A)=\{(-1,6,2,9),(0,21,7,5)\}$
And colspace $A= \begin{bmatrix}
-1 & 6 & 2 & 0\\
3 & 3 &1 & 5 \\
7 & 21 & 7 & 15
\end{bmatrix} \approx \begin{bmatrix}
-1 & 0 & 0 & 0\\
3 & 21 & 7 & 5 \\
7 & 63 & 21 & 15
\end{bmatrix} \approx \begin{bmatrix}
-1 & 0 & 0 & 0\\
3 & 21 & 0 & 0\\
7 & 63 & 0 & 0
\end{bmatrix} \\
colspace (A)=\{(-1,3,7),(0,21,63)\}$
Thus the nullspace $(A)$ can be written as $Ax=0 \rightarrow \begin{bmatrix}
-1 & 6 & 2 & 0\\
0 & 21 & 7 & 5 \\
0 & 0 & 0 & 0
\end{bmatrix} \begin{bmatrix}
x\\
y \\
z\\
w
\end{bmatrix}= \begin{bmatrix}
0\\
0 \\
0\\0
\end{bmatrix} \\
\rightarrow 21y+7z+5w=0 \rightarrow y=-\frac{1}{3}y-\frac{5}{21}z\\
\rightarrow -x+6y+2z=0 \rightarrow x=6y+2z=-\frac{10}{7}z$
Thus, the null space $A$ is $\{(-\frac{1}{3}y-\frac{5}{21}z,-\frac{10}{7}z)\}$
Since $dim(nullspace (A))=2$ we have $z=0, w=1 \\
z=1, w=0$
Consequently, the null space of the given matrix $A$ is $\{(-\frac{10}{7},-\frac{5}{21},0,1),(0,-\frac{1}{3},1,0)\}$
