Romanian Mathematical Olympiad

a) We shall use induction on $n$. As $x_2=a\cdot 1+b\cdot 0=a$, we have $$A=\left[\begin{array}{cc}a& b\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}{x}_2& b{x}_1\\ x_1& b{x}_0\end{array}\right],$$ so the property is true for $n=1$. Consider now that the formula is true for some $n\ge 1$. Recurrently we then get $$A^{n+1}=A^n\cdot A=\left[\begin{array}{cc}{x}_{n+1}& b{x}_n\\ x_n& b{x}_{n-1}\end{array}\right]\cdot \left[\begin{array}{cc}a& b\\ 1& 0\end{array}\right]=\left[\begin{array}{cc}{x}_{n+2}& b{x}_{n+1}\\ x_{n+1}& b{x}_n\end{array}\right],$$ ending the induction.