MDoubt No 000019 - GATE

GATE [EC, 2017, 1 mark]
Question : Consider the $5\times 5$ matrix $$A=\begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 5 & 1 & 2 & 3 & 4\\ 4 & 5 & 1 & 2 & 3\\ 3 & 4 & 5 & 1 & 2\\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$$ 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) $25$

Solution : The characteristic equation is $|A-\lambda I|=0$
$\implies\begin{vmatrix} 1-\lambda & 2 & 3 & 4 & 5\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda & 2\\ 2 & 3 & 4 & 5 & 1-\lambda \end{vmatrix}=0$
$R_1\to R_1+R_2+R_3+R_4+R_5$
$\implies\begin{vmatrix} 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda & 2\\ 2 & 3 & 4 & 5 & 1-\lambda \end{vmatrix}=0$
taking out $15-\lambda$ from $R_1$
$\implies[15-\lambda]\begin{vmatrix} 1 & 1 & 1 & 1 & 1\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda & 2\\ 2 & 3 & 4 & 5 & 1-\lambda \end{vmatrix}=0$
$\implies 15-\lambda=0$
$\implies\lambda=15$
CORRECT ANSWER : C