Eigen values of the matrix \[\left[ {\begin{array}{*{20}{c}} 3&{ – 1}&{ – 1} \\ { – 1}&3&{ – 1} \\ { – 1}&{ – 1}&3 \end{array}} \right]\] are A. 1, 1, 1 B. 1, 1, 2 C. 1, 4, 4 D. 1, 2, 4

The correct answer is $\boxed{\text{A) }1, 1, 1}$.

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

$$\lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.

The characteristic polynomial of the matrix $A$ is given by:

$$p(x) = |A – xI| = \det \left[ {\begin{array}{*{20}{c}} 3&{ – 1}&{ – 1} \ { – 1}&3&{ – 1} \ { – 1}&{ – 1}&3 \end{array}} \right] = x^3 – 9x^2 + 6x – 3$$

To find the eigenvalues of $A$, we need to solve the equation $p(x) = 0$.

Solving $p(x) = 0$, we get the following eigenvalues:

$$\lambda = 1, 1, 1$$

Therefore, the eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$.

Here is a brief explanation of each option:

• Option A: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 1$ into the characteristic polynomial $p(x)$.
• Option B: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 2$ into the characteristic polynomial $p(x)$.
• Option C: The eigenvalues of the matrix $A$ are $\boxed{\text{A) }1, 1, 1}$. This can be verified by substituting $x = 4$ into the characteristic polynomial $p(x)$.