Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Digital systems, linear algebra, introduction to formal proof / 4 / d / Solution

The diagonalization gives $A=SD{S}^{-1}$ and hence $A^{n-1}=S{D}^{n-1}{S}^{-1}$: the intervening factors $S^{-1}S$ cancel. The diagonal matrix power has entries ${\alpha }^{n-1},{\beta }^{n-1}$. Since $1-\beta =\alpha$ and $\alpha -1=-\beta$,Reading the second component gives the Binet formulaThe derivation includes $n=1$, and the expression gives $U_1=U_2=1$, consistent with the specified indexing.