Question

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

• A. 1,2
• B. -1,2
• C. -1,-2
• D. 1,-2

Answer 1

Answer: (B)

-1,2

Answer 2

We have,
$\begin{vmatrix}x-1 & 1 & 1 \\\ 1 & x-1 & 1 \\\ 1 & 1 & x-1\end{vmatrix}=0$
On applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$, we get
$\begin{vmatrix}x+1 & 1 & 1 \\\ x+1 & x-1 & 1 \\\ x+1 & 1 & x-1\end{vmatrix}=0$
On taking $(x+1)$ common from $C_{1}$, we get
$(x+1)\begin{vmatrix}1 & 1 & 1 \\\ 1 & x-1 & 1 \\\ 1 & 1 & x-1\end{vmatrix}=0$
On applying, $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$, we get
$\Rightarrow(x+1)\begin{vmatrix}0 & 2-x & 0 \\\ 0 & x-2 & 2-x \\\ 1 & 1 & x-1\end{vmatrix}=0$
$\Rightarrow (x+1) .1\left[(2-x)^{2}-0\right]=0$
$\Rightarrow (x+1)(2-x)^{2}=0$
$\Rightarrow x=-1,2$