Answer
a) $x$ is an eigenvector of $A$.
b) $x$ is not an eigenvector of $A$.
c) $x$ is an eigenvector of $A$.
d) $x$ is an eigenvector of $A$.
Work Step by Step
Let $A$ be any $n\times n$ matrix. Let $\lambda$ be some scalar for which there exists a non zero vector $x$ such that $Ax=\lambda x$. Then $\lambda$ is called an eigenvalue of $A$ and $x$ is called the corresponding eigenvalue.
Given $A= \left[ {\begin{array}{*{20}{c}}
{-1}&-1&1\\
-2&0&-2\\
3&-3&1
\end{array}} \right]$.
a) Given $x = \left[ {\begin{array}{*{20}{c}}
2\\
-4\\
6
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-1}&-1&1\\
-2&0&-2\\
3&-3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
-4\\
6
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{8}\\
{-16}\\
24
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{8}\\
{-16}\\
24
\end{array}} \right] =4 \left[ {\begin{array}{*{20}{c}}
2\\
-4\\
6
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
b) Given $x = \left[ {\begin{array}{*{20}{c}}
2\\
0\\
6
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-1}&-1&1\\
-2&0&-2\\
3&-3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
0\\
6
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{4}\\
{-16}\\
12
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{4}\\
{-16}\\
12
\end{array}} \right] \neq \lambda\left[ {\begin{array}{*{20}{c}}
2\\
0\\
6
\end{array}} \right]$ for any scalar $\lambda \in R$
Therefore, $x$ is an eigenvector of $A$.
c) Given $x = \left[ {\begin{array}{*{20}{c}}
2\\
2\\
0
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-1}&-1&1\\
-2&0&-2\\
3&-3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
2\\
0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{-4}\\
{-4}\\
0
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{-4}\\
{-4}\\
0
\end{array}} \right] ={-2} \left[ {\begin{array}{*{20}{c}}
2\\
2\\
0
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
d) Given $x = \left[ {\begin{array}{*{20}{c}}
-1\\
0\\
1
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-1}&-1&1\\
-2&0&-2\\
3&-3&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
-1\\
0\\
1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{2}\\
{0}\\
-2
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{2}\\
{0}\\
-2
\end{array}} \right] ={-2} \left[ {\begin{array}{*{20}{c}}
-1\\
0\\
1
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
