Answer
$\left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right]$ is an eigenvector of $A$ for the eigenvalue 0.
Work Step by Step
Given $$ A=\left[\begin{array}{ccc}
3 & 7 & 9 \\
-4 & -5 & 1 \\
2 & 4 & 4
\end{array}\right],\ \ \ \ \ \mathbf{v}=\left[\begin{array}{c}
4 \\
-3 \\
1
\end{array}\right]$$
Since
\begin{align*}
A \mathbf{v}&=\left[\begin{array}{ccc}
3 & 7 & 9 \\
-4 & -5 & 1 \\
2 & 4 & 4
\end{array}\right]\left[\begin{array}{c}
4 \\
-3 \\
1
\end{array}\right]\\
&=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] =0\left[\begin{array}{c}
4 \\
-3 \\
1
\end{array}\right]
\end{align*}
then $A \mathbf{v}=0 \mathbf{v}$, hence $\left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right]$ is an eigenvector of $A$ for the eigenvalue 0.
