Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
1 & -2 & 1\\ 4 & -7 & -2 \\ -1 & 3 &4
\end{bmatrix}$
Since $x,y,z \in nullspace (A)$ we obtain
$Ax=0\\\begin{bmatrix}
1 & -2 & 1\\ 4 & -7 & -2 \\ -1 & 3 &4
\end{bmatrix}\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}=0\\
\begin{bmatrix}
x -2y +z\\ 4x -7y -2z \\ -x+3y+4z
\end{bmatrix}=0\\
\rightarrow x-2y+z=0\\
4x-7y-2z=0\\
-x+3y+4z=0\\
\rightarrow x-2y+z+(-x+3y+4z)=y+5z=0\\
2(x-2y+z)+4x-7y-2z=6x-11y=0\\
2(4x-7y-2z)+(-x+3y+4z)=7x-11y=0$
We have the system:
$y+5z=0\\
6x-11y=0\\
7x-11y=0$
then $7x-11y-(6x-11y)=0 \rightarrow x=0\\
6(0)-11y=0\rightarrow y=0\\
0+5z=0 \rightarrow z=0$
Hence, nullspace $(A) =\{(0,0,0)\}$
