# The roots of the equation $\begin{vmatrix}x-1&1&1\\1&x-1&1\\1&1&x-1\end{vmatrix}=0$ are

$\begin{array}{1 1}(A)\;1,2&(B)\;-1,2\\(C)\;1,-2&(D)\;-1,-2\end{array}$

Given :
$\begin{vmatrix}x-1&1&1\\1&x-1&1\\1&1&x-1\end{vmatrix}=0$
Operating $C_1=C_1+C_2+C_3$
$\begin{vmatrix}x+1&1&1\\x+1&x-1&1\\x+1&1&x-1\end{vmatrix}=0$
$\Rightarrow (x+1)\begin{vmatrix}1& 1&1\\1&x-1&1\\1&1&x-1\end{vmatrix}=0$
Apply $R_2-R_1$ and $R_3-R_1$ and get
$(x+1)\begin{vmatrix}1&1&1\\0&x-2&0\\0&0&x-2\end{vmatrix}=0$
$\Rightarrow (x+1)(x-2)^2=0$
$x=-1,2$
Hence (B) is the correct answer.