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The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{x - 1}&1&1\\1&{x - 1}&1\\1&1&{x - 1}\end{array}\,} \right| = 0$ are
$1, 2$
$-1, 2$
$1, -2$
$-1, -2$
Solution
(b) We have $\left| {\,\begin{array}{*{20}{c}}{x – 1}&1&1\\1&{x – 1}&1\\1&1&{x – 1}\end{array}\,} \right|\, = 0$
$ \Rightarrow $$\left| {\,\begin{array}{*{20}{c}}{x + 1}&1&1\\{x + 1}&{x – 1}&1\\{x + 1}&1&{x – 1}\end{array}\,} \right|\, = 0$,
{Applying ${C_1} \to {C_1} + {C_2} + {C_3}$}
$ \Rightarrow $$(x + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{x – 1}&1\\1&1&{x – 1}\end{array}\,} \right|$= 0
$ \Rightarrow $ $(x + 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\0&{x – 2}&0\\0&0&{x – 2}\end{array}\,} \right| = 0$
{Applying ${R_2} \to {R_2} – {R_1},\,{R_3} \to {R_3} – {R_1}$}
$\Rightarrow $ $(x+1) (x-2)^2 = 0 => x =-1,2. $
