The minimum and the maximum eigen values of the matrix \[\left[ {\begin{array}{*{20}{c}} 1&1&3 \\ 1&5&1 \\ 3&1&1 \end{array}} \right]\] are -2 and 6, respectively. What is the other eigen value? A. 5 B. 3 C. 1 D. -1

The correct answer is $\boxed{3}$.

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:

$$p(x) = |A – xI|$$

In this case, we have:

$$p(x) = | \left[ {\begin{array}{*{20}{c}} 1&1&3 \ 1&5&1 \ 3&1&1 \end{array}} \right] – xI| = x^3 – 6x^2 + 10x – 1$$

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

Solving this equation, we find that the eigenvalues of $A$ are $-2$, $3$, and $6$.

Therefore, the other eigenvalue of the matrix $A$ is $\boxed{3}$.

Here is a brief explanation of each option:

• Option A: $5$ is not an eigenvalue of $A$. This can be verified by substituting $x = 5$ into the characteristic polynomial $p(x)$ and evaluating.
• Option B: $3$ is an eigenvalue of $A$. This can be verified by substituting $x = 3$ into the characteristic polynomial $p(x)$ and evaluating.
• Option C: $1$ is an eigenvalue of $A$. This can be verified by substituting $x = 1$ into the characteristic polynomial $p(x)$ and evaluating.
• Option D: $-1$ is not an eigenvalue of $A$. This can be verified by substituting $x = -1$ into the characteristic polynomial $p(x)$ and evaluating.