The matrix \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} {\frac{3}{2}}&0&{\frac{1}{2}} \\ 0&{ – 1}&0 \\ {\frac{1}{2}}&0&{\frac{3}{2}} \end{array}} \right]\] has three distinct eigen values and one of its eigen vectors is \[\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right].\] Which one of the following can be another eigen vector of A? A. \[\left[ {\begin{array}{*{20}{c}} 0 \\ 0 \\ { – 1} \end{array}} \right]\] B. \[\left[ {\begin{array}{*{20}{c}} { – 1} \\ 0 \\ 0 \end{array}} \right]\] C. \[\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ { – 1} \end{array}} \right]\] D. \[\left[ {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 1 \end{array}} \right]\]

\]” option2=”\[\left[ {\begin{array}{*{20}{c}} { – 1} \\ 0 \\ 0 \end{array}} \right]\]” option3=”\[\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ { – 1} \end{array}} \right]\]” option4=”\[\left[ {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ 1 \end{array}} \right]\]” correct=”option1″]

The correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 1 \ 0 \ { – 1} \end{array}} \right]}$.

An eigenvector of a matrix $A$ is a nonzero vector $v$ such that $Av=\lambda v$ for some scalar $\lambda$. In other words, an eigenvector is a vector that is scaled by a constant when it is multiplied by $A$.

The matrix $A$ has three distinct eigenvalues, which means that it has three linearly independent eigenvectors. One of these eigenvectors is $\left[ {\begin{array}{*{20}{c}} 1 \ 0 \ 1 \end{array}} \right]$.

To find another eigenvector of $A$, we can use the following procedure:

1. Let $v$ be a vector that is not a scalar multiple of $\left[ {\begin{array}{*{20}{c}} 1 \ 0 \ 1 \end{array}} \right]$.
2. Multiply $v$ by $A$ to get $Av$.
3. Solve the equation $Av=\lambda v$ for $\lambda$.
4. If $\lambda$ is an eigenvalue of $A$, then $v$ is an eigenvector of $A$ corresponding to $\lambda$.

In this case, we can let $v=\left[ {\begin{array}{*{20}{c}} x \ y \ z \end{array}} \right]$. Multiplying $v$ by $A$, we get

[Av=\left[ {\begin{array}{{20}{c}} {\frac{3}{2}}x&0&{\frac{1}{2}}y \ 0&{ – 1}y&0 \ {\frac{1}{2}}z&0&{\frac{3}{2}}z \end{array}} \right]\left[ {\begin{array}{{20}{c}} x \ y \ z \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} {\frac{3}{2}}x-\frac{1}{2}yz \ -y \ \frac{3}{2}z-\frac{1}{2}xy \end{array}} \right]]

Solving the equation $Av=\lambda v$ for $\lambda$, we get

[\left[ {\begin{array}{{20}{c}} {\frac{3}{2}}x-\frac{1}{2}yz \ -y \ \frac{3}{2}z-\frac{1}{2}xy \end{array}} \right]=\lambda \left[ {\begin{array}{{20}{c}} x \ y \ z \end{array}} \right]]

This gives us the system of equations

\begin{align}
{\frac{3}{2}}x-\frac{1}{2}yz &= \lambda x \
-y &= \lambda y \
\frac{3}{2}z-\frac{1}{2}xy &= \lambda z
\end{align}

Solving this system of equations, we get

\begin{align}
x &= \frac{2\lambda}{3+z} \
y &= \frac{2\lambda}{3} \
z &= \frac{2\lambda}{3-x}
\end{align}

Therefore, an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ is

[\left[ {\begin{array}{*{20}{c}} \frac{2\lambda}{3+z} \ \frac{2\lambda}{3} \ \frac{2\lambda}{3-x} \end{array}} \right]]

For $\lambda=1$, this gives us the eigenvector $\left[ {\begin{array}{*{20}{c}} 1 \ 0 \ -1 \end{array}} \right]$.