Explanation

(b)
For singular matrix,
$\begin{bmatrix}
- x & x & 2 \\
2 & x & - x \\
x & - 2 & - x
\end{bmatrix} = 0$
Applying $C_{2} \rightarrow C_{2} + C_{1},\ C_{3} \rightarrow C_{3} + C_{1}$
$\Rightarrow \ \left| \begin{matrix}
- x & 0 & 2 - x \\
2 & 2 + x & 2 - x \\
x & x - 2 & 0
\end{matrix} \right| = 0$
$\Rightarrow (2 - x)\ \left| \begin{matrix}
- x & 0 & 1 \\
2 & 2 + x & 1 \\
x & x - 2 & 0
\end{matrix} \right| = 0$
Applying $R_{2} \rightarrow R_{2} - R_{1}$
$\Rightarrow (2 - x)\left| \begin{matrix}
- x & 0 & 1 \\
2 + x & 2 + x & 0 \\
x & x - 2 & 0
\end{matrix} \right| = 0$
$\Rightarrow$($2 - x)(2 + x)\left| \begin{matrix}
- x & 0 & 1 \\
1 & 1 & 0 \\
x & x - 2 & 0
\end{matrix} \right| = 0$
$\Rightarrow (2 - x)(2 + x)(x - 2 - x) = 0$
$\Rightarrow x = 2,\ - 2$
$\therefore Given\ matrix\ is\ non - singular\ for\ all\ x\ other\ than\ 2\ and - 2$.