Step 1
Given Matrix \[A = \left[ {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right]\]Eigenvector:\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right]\]We have to find the corresponding eigenvalue for the above eigenvector.
An eigenvector of an $n \times n$ matrix A is a nonzero vector x such that $A{\bf{x}} = \lambda {\bf{x}}$ for some scalar $\lambda $. A scalar $\lambda $ is called an eigenvalue of A if there is a nontrivial solution $\bf{x}$ of $A{\bf{x}} = \lambda {\bf{x}}$; such an $\bf{x}$ is called an eigenvector corresponding to $\lambda $.



Step 2: The equation form
Write the equation to find eigenvalues.\[\begin{array}{l}A{\bf{x}} = \lambda {\bf{x}}\\\left[ {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{4 \times 3 - 7 \times 3 + 9 \times 1}\\{ - 4 \times 4 + 5 \times 3 + 1 \times 1}\\{2 \times 4 - 3 \times 4 + 4 \times 1}\end{array}} \right] = \lambda \left[ {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4\lambda }\\{ - 3\lambda }\\\lambda \end{array}} \right]\end{array}\]

ANSWER
As from the above matrix form of equation that the only possible solution is when $\lambda = 0$.

$\lambda = 0$ is an eigenvalue for the matrix and corresponding eigenvector.