Let $a_n, b_n$ be sequences of positive real numbers which satisfy $a_0 = b_0$ and the recursive matrix relation $\big(\begin{smallmatrix}a_n \\b_n\end{smallmatrix}\big)=\big(\begin{smallmatrix} 1 & y \\0 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix}a_{n-1} \\b_{n-1}\end{smallmatrix}\big).$ What is the simplest non-recursive form of the ratio $\frac{b_n}{a_n}?$

Evaluate $\big(\begin{smallmatrix}1 & y \\0 & 1\end{smallmatrix}\big)\big(\begin{smallmatrix}u \\v\end{smallmatrix}\big).$

Try to express the relationship between $\big(\begin{smallmatrix}a_n \\ b_n\end{smallmatrix}\big)$ and $\big(\begin{smallmatrix}a_0 \\ b_0\end{smallmatrix}\big)$ as a non-recursive matrix equation.

Simplify ${\big(\begin{smallmatrix}1 & y \\0 & 1\end{smallmatrix}\big)}^n,$ by trying small values of $n$ or otherwise.

We can unwind this matrix recurrence in the same way as a regular recurrence, $\big(\begin{smallmatrix} a_n \\ b_n \end{smallmatrix}\big) = \big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big) \big(\begin{smallmatrix} a_{n-1} \\ b_{n-1} \end{smallmatrix}\big)$ $={\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^2 \big(\begin{smallmatrix} a_{n-2} \\ b_{n-2} \end{smallmatrix}\big)$ $=\ldots=$ ${\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^n \big(\begin{smallmatrix} a_{0} \\ b_{0} \end{smallmatrix}\big).$ Using induction, we can show that ${\big(\begin{smallmatrix} 1 & y \\ 0 & 1 \end{smallmatrix}\big)}^n$ $={\big(\begin{smallmatrix} 1 & ny \\ 0 & 1 \end{smallmatrix}\big)}.$ Hence, $\big(\begin{smallmatrix} a_n \\ b_n \end{smallmatrix}\big)$ $={\big(\begin{smallmatrix} 1 & ny \\ 0 & 1 \end{smallmatrix}\big)} \big(\begin{smallmatrix} a_{0} \\ b_{0} \end{smallmatrix}\big)$ $=\big(\begin{smallmatrix} a_{0} + nyb_{0} \\ b_{0} \end{smallmatrix}\big).$ Therefore the ratio $\frac{b_n}{a_n}$ is $\frac{b_0}{a_0 + nyb_0} = \frac{1}{1+ny}.$