Given $\mathrm{A}=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]$ satisfies Aⁿ = A^n–2 + A² – 1 for n ≥ 3
⇒ A⁴ = A² + A² – I = 2A² – I
⇒ A⁶ = A⁴ + A² – I = 3A² – 2I
similarly
A⁵⁰ = 25A² – 24I
${\mathrm{A}}^2=\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]$
$\therefore {\mathrm{A}}^{50}=\left[\begin{array}{ccc}25& 0& 0\\ 25& 25& 0\\ 25& 0& 25\end{array}\right]-\left[\begin{array}{ccc}24& 0& 0\\ 0& 24& 0\\ 0& 0& 24\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 25& 1& 0\\ 25& 0& 1\end{array}\right]$
$\Rightarrow {\mathrm{A}}^{50}=\left[\begin{array}{ccc}1& 0& 0\\ 25& 1& 0\\ 25& 0& 1\end{array}\right]$
∴ |A⁵⁰| = 1
Hence, option B.