Consider the 5 × 5 matrix \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&1&2&3&4 \\ 4&5&1&2&3 \\ 3&4&5&1&2 \\ 2&3&4&5&1 \end{array}} \right]\] It is given that A has only one real eigen value. Then the real eigen value of A is A. -2.5 B. 0 C. 15 D. 25

The correct answer is $\boxed{0}$.

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:

$$p(x) = |xI – A| = x^5 – 15x^4 + 75x^3 – 125x^2 + 75x – 15$$

We can factor the characteristic polynomial as follows:

$$p(x) = (x – 1)(x – 5)(x – 3)(x – 2)(x – 1)$$

This means that the eigenvalues of $A$ are $1$, $5$, $3$, $2$, and $1$.

Since $A$ has only one real eigenvalue, the real eigenvalue of $A$ must be $1$.

Here is a brief explanation of each option:

• Option A: $-2.5$ is not an eigenvalue of $A$. This can be verified by substituting $x = -2.5$ into the characteristic polynomial $p(x)$.
• Option B: $0$ is an eigenvalue of $A$. This can be verified by substituting $x = 0$ into the characteristic polynomial $p(x)$.
• Option C: $15$ is not an eigenvalue of $A$. This can be verified by substituting $x = 15$ into the characteristic polynomial $p(x)$.
• Option D: $25$ is not an eigenvalue of $A$. This can be verified by substituting $x = 25$ into the characteristic polynomial $p(x)$.