Given the matrix \[\left[ {\begin{array}{*{20}{c}} { – 4}&2 \\ 4&3 \end{array}} \right],\] the eigen vector is A. \[\left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]\] B. \[\left[ {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right]\] C. \[\left[ {\begin{array}{*{20}{c}} 2 \\ { – 1} \end{array}} \right]\] D. \[\left[ {\begin{array}{*{20}{c}} { – 1} \\ 2 \end{array}} \right]\]

\]” option2=”\[\left[ {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right]\]” option3=”\[\left[ {\begin{array}{*{20}{c}} 2 \\ { – 1} \end{array}} \right]\]” option4=”\[\left[ {\begin{array}{*{20}{c}} { – 1} \\ 2 \end{array}} \right]\]” correct=”option3″]

The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 2 \ { – 1} \end{array}} \right]}$.

To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:

$$\lambda v = A v$$

where $\lambda$ is the eigenvalue, $v$ is the eigenvector, and $A$ is the matrix.

In this case, we have the matrix $A = \left[ {\begin{array}{*{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right]$.

To find the eigenvalues, we can solve the equation $Av = \lambda v$.

We can do this by first finding the characteristic polynomial of $A$, which is given by:

$$| A – \lambda I |$$

In this case, the characteristic polynomial is:

$$| \left[ {\begin{array}{{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right] – \lambda \left[ {\begin{array}{{20}{c}} 1&0 \ 0&1 \end{array}} \right] | = \left[ {\begin{array}{*{20}{c}} { – 4}-\lambda &2 \ 4&3-\lambda \end{array}} \right]$$

We can then solve the equation $| A – \lambda I | = 0$ to find the eigenvalues.

In this case, the eigenvalues are $\lambda = 2$ and $\lambda = 5$.

Once we have found the eigenvalues, we can find the

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