The matrix \[{\text{M}} = \left[ {\begin{array}{*{20}{c}} { – 2}&2&{ – 3} \\ 2&1&{ – 6} \\ { – 1}&{ – 2}&0 \end{array}} \right]\] has eigen values -3, -3, 5. An eigen vector corresponding to the eigen value 5 is [1 2 -1]T. One of the eigen vectors of the matrix M3 is A. [1 8 -1]T B. [1 2 -1]T C. \[{\left[ {\begin{array}{*{20}{c}} 1&{\sqrt[3]{2}}&{ – 1} \end{array}} \right]^{\text{T}}}\] D. [1 1 -1]T

T” option2=”[1 2 -1]T” option3=”\[{\left[ {\begin{array}{*{20}{c}} 1&{\sqrt[3]{2}}&{ – 1} \end{array}} \right]^{\text{T}}}\]” option4=”[1 1 -1]T” correct=”option1″]

The correct answer is $\boxed{\text{C}}$.

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

We are given that the matrix $M$ has eigenvalues $-3$, $-3$, and $5$. We are also given that an eigenvector corresponding to the eigenvalue $5$ is $[1 2 -1]^T$. This means that $M[1 2 -1]^T=5[1 2 -1]^T$. We can rewrite this equation as $M-5I[1 2 -1]^T=0$.

To find the other eigenvectors of $M$, we can use the following formula:

$$v_2=\frac{1}{\|M-5I\|}\left(M-5I\right)v_1$$

where $v_1$ is an eigenvector of $M$ corresponding to the eigenvalue $5$ and $\|M-5I\|$ is the determinant of the matrix $M-5I$.

In this case, $\|M-5I\|=-2$. Therefore, the other eigenvector of $M$ is:

$$v_2=\frac{1}{-2}\left(M-5I\right)[1 2 -1]^T=\left[{\begin{array}{*{20}{c}} 1&{\sqrt[3]{2}}&{ – 1} \end{array}} \right]^T$$

Therefore, one of the eigenvectors of the matrix $M^3$ is $\boxed{\left[ {\begin{array}{*{20}{c}} 1&{\sqrt[3]{2}}&{ – 1} \end{array}} \right]^T}$.