Explanation:
(D) : Given that,
\(A=\left[\begin{array}{ll}
x & 1 \\ 1 & 0 \end{array}\right]\)
We know that,
\(\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}\)
Now, \(|A|=0-1=-1\)
And \(\quad \operatorname{adj} A=\left[\begin{array}{cc}0 & -1 \\ -1 & \mathrm{x}\end{array}\right]\)
\(\therefore \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}=\frac{1}{-1}\left[\begin{array}{cc}0 & -1 \\ -1 & \mathrm{x}\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ 1 & -\mathrm{x}\end{array}\right]\)
According to question,
\(\mathrm{A}=\mathrm{A}^{-1}\)
\(\left[\begin{array}{ll}
\mathrm{x} & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{cc} 0 & 1 \\ 1 & -\mathrm{x} \end{array}\right]\)
On comparing corresponding elements on both side, we get-
\(\mathrm{x}=0\)