Answer
a) $x$ is not an eigenvector of $A$.
b) $x$ is an eigenvector of $A$.
c) $x$ is an eigenvector of $A$.
d) $x$ is not 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 $A= \left[ {\begin{array}{*{20}{c}}
7&2\\
2&4
\end{array}} \right]$.
a) Given $x = \left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
7&2\\
2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{11}\\
{10}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{11}\\
{10}
\end{array}} \right] \ne \lambda\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]$ for any scalar $\lambda \in R$
Therefore, $x$ is not an eigenvector of $A$.
b) Given $x = \left[ {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
7&2\\
2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{16}\\
{8}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{16}\\
{8}
\end{array}} \right] = 8\left[ {\begin{array}{*{20}{c}}
2\\
1
\end{array}} \right]=8x$.
Therefore, $x$ is an eigenvector of $A$.
c) Given $x = \left[ {\begin{array}{*{20}{c}}
1\\
{-2}
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
7&2\\
2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
{-2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{3}\\
{-6}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{3}\\
{-6}
\end{array}} \right] =3 \left[ {\begin{array}{*{20}{c}}
3\\
{-6}
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
d) Given $x = \left[ {\begin{array}{*{20}{c}}
{-1}\\
0
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
7&2\\
2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{-1}\\
0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{-7}\\
{-2}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{-7}\\
{-2}
\end{array}} \right] \ne \lambda\left[ {\begin{array}{*{20}{c}}
{-1}\\
0
\end{array}} \right]$ for any scalar $\lambda \in R$
Therefore, $x$ is not an eigenvector of $A$.
