The correct answer is (c) 3

To elaborate: The eigenvalue, L of a matrix is equal to the root (factor) of the equation |K-LI|=0.

Let the given matrix be denoted by K then K-LI = \(\begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix} – L \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix}\)

= \(\begin{pmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{pmatrix}\)

|K-LI| = \(\begin{vmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{vmatrix}\)

= (1-L)((1-L)^2-1)-1(-L)+1(L)

= (1-L)(L^2-2L) + 2L

= -L^3 + 3L^2

= -L^2 (L-3).

Given -L^2 (L-3) = 0

On simplification L = 0, 0 and 3.