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(B) Given the determinant equation: $\left| \begin{array}{ccc} x-1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying the column o

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$-1, 2$

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(B) Given the determinant equation: $\left| \begin{array}{ccc} x-1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying the column operation $C_1 	o C_1 + C_2 + C_3$:$\left| \begin{array}{ccc} x+1 & 1 & 1 \ x+1 & x-1 & 1 \ x+1 & 1 & x-1 \end{array} ight| = 0$Taking $(x+1)$ common from $C_1$:$(x+1) \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$(x+1) \left| \begin{array}{ccc} 1 & 1 & 1 \ 0 & x-2 & 0 \ 0 & 0 & x-2 \end{array} ight| = 0$Expanding along $C_1$:$(x+1) [1 \cdot (x-2)(x-2) - 0 + 0] = 0$$(x+1)(x-2)^2 = 0$Thus,the roots are $x = -1$ and $x = 2$.

The roots of the equation $\left| \begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array} \right| = 0$ are

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