how to use matrix to prove this identity?

Question

if $a_{n},b_{n}$ such $a_{0}=b_{0}=1$ $$\begin{cases}a_{n}=5a_{n-1}+7b_{n-1}\\ b_{n}=7a_{n-1}+10b_{n-1},\forall n=1,2,3,\cdots \end{cases}$$

show that $$a_{m+n}+b_{m+n}=a_{m}a_{n}+b_{m}b_{n}$$ It's an interesting identity, and I've proved it with mathematical induction, but I feel like it's more obvious with a matrix, but I don't, so can someone please prove it with a matrix? Thank you $$\begin{bmatrix} a_{n}\\ b_{n}\end{bmatrix}=\begin{bmatrix} 5&7\\ 7&10\end{bmatrix}\cdot \begin{bmatrix} a_{n-1}\\ b_{n-1}\end{bmatrix}$$

Answer 1

$$ \begin{aligned} \begin{Bmatrix} a_{0} \\ b_{0} \end{Bmatrix}^{\top} \begin{Bmatrix} a_{m+n}\\ b_{m+n} \end{Bmatrix} &= \begin{Bmatrix} a_{0} \\ b_{0} \end{Bmatrix}^{\top} \begin{bmatrix} 5&7\\7&10 \end{bmatrix}^{m+n} \begin{Bmatrix} a_{0}\\ b_{0} \end{Bmatrix}\\ &= \begin{Bmatrix} a_{0} \\ b_{0} \end{Bmatrix}^{\top} \begin{bmatrix} 5&7\\7&10 \end{bmatrix}^{m} \begin{bmatrix} 5&7\\7&10 \end{bmatrix}^{n} \begin{Bmatrix} a_{0}\\ b_{0} \end{Bmatrix}\\ &= \left(\begin{bmatrix} 5&7\\7&10 \end{bmatrix}^{m} \begin{Bmatrix} a_{0}\\ b_{0} \end{Bmatrix} \right)^{T} \begin{bmatrix} 5&7\\7&10 \end{bmatrix}^{n} \begin{Bmatrix} a_{0}\\ b_{0} \end{Bmatrix}\\ &=\begin{Bmatrix} a_{m} \\ b_{m} \end{Bmatrix}^{\top} \begin{Bmatrix} a_{n}\\b_{n} \end{Bmatrix} \end{aligned} $$

This is possible because transposing Your square matrix does not change it.