The value(s) of x, for which the matrix A=≤ft[ […] ] is singular, is/…

$A=\left[\begin{array}{ccc}x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right]$

we need to find x for |A| = 0

$|A|=\left|\begin{array}{ccc}x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$

⇒ Using operation R₁ → R₁ + R₂ + R₃

$|A|=\left|\begin{array}{ccc}x+1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$

⇒  So  $|A|=(x+1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1\end{array}\right|$

⇒ Using operation R₂ → R₂ - R₁

R₃ → R₃ - R₁

$(x+1)\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & x-2 & 1 \\ 1 & 1 & x-2\end{array}\right|$

So expanding along R₁

So |A| = (x + 1)(x - 2)(x - 2) = 0

⇒ x = -1, 2  for  |A| = 0 (singularity)