Answer
See answer below
Work Step by Step
We are given $A=\begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
6 & 5& 2 \\
-2 & 2 & 10
\end{bmatrix}$
We obtain rowspace $A=\begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
6 & 5& 2 \\
-2 & 2 & 10
\end{bmatrix} \approx \begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
0 & 11 & 32 \\
0 & -4 & -17
\end{bmatrix} \approx \begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
0 & 0 & 277 \\
0 & 0 & 59
\end{bmatrix} \approx\begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
0 & 0 & 277 \\
0 & 0 & 0
\end{bmatrix} \\
rowspace (A)=\{(-4,0,3),(0,10,13),(0,0,277)\}$
And colspace $A= \begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
0 & 0 & 277 \\
0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
-4 & 0& 0\\
0 & 10 & 39 \\
6 & 5 & 26 \\
-2 & 2 & 34
\end{bmatrix} \approx \begin{bmatrix}
-4 & 0& 0\\
0 & 10 & 0\\
6 & 5 & -6.5 \\
-2 & 2 & -262
\end{bmatrix} \\
colspace (A)=\{(-4,0,6,-2),(0,10,5,2),(0,0,-6.5,-262)\}$
Thus the nullspace $(A)$ can be written as $Ax=0 \rightarrow \begin{bmatrix}
-4 & 0& 3\\
0 & 10 & 13 \\
0 & 0 & 277 \\
0 & 0 & 0
\end{bmatrix} \begin{bmatrix}
x\\
y \\
z
\end{bmatrix}= \begin{bmatrix}
0\\
0 \\
0\\0
\end{bmatrix} \\
\rightarrow x=y=z=0$
Consequently, the null space of the given matrix $A$ is $\{(0,0,0)\}$
