Let $P=\begin{bmatrix}1-a&a\\b&1-b\end{bmatrix}$, with $0<a,b<1$. Show that $$P^n=\frac{1}{a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix}+\frac{(1-a-b)^n}{a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}$$
I think it's possible to prove using induction principle, but I do not know if it's the best way
Let $n=1$ $$P^1=\frac{1}{a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix}+\frac{(1-a-b)^1}{a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}=\frac{1}{a+b}\left(\begin{bmatrix}b&a\\b&a\end{bmatrix}+(1-a-b)^1\begin{bmatrix}a&-a\\-b&b\end{bmatrix}\right)=\begin{bmatrix}1-a&a\\b&1-b\end{bmatrix}$$
Assuming that is true for $n=k$ $$P^k=\frac{1}{a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix}+\frac{(1-a-b)^k}{a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}$$
Now I need to show that is true for $n=k+1$ $$P^{k+1}=\frac{1}{a+b}\begin{bmatrix}b&a\\b&a\end{bmatrix}+\frac{(1-a-b)^{k+1}}{a+b}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}=\frac{1}{a+b}\left(\begin{bmatrix}b&a\\b&a\end{bmatrix}+(1-a-b)^{k+1}\begin{bmatrix}a&-a\\-b&b\end{bmatrix}\right)=\frac{1}{a+b}\left(\begin{bmatrix}b&a\\b&a\end{bmatrix}+(1-a-b)^{k}(1-a-b)\begin{bmatrix}a&-a\\-b&b\end{bmatrix}\right)$$
but now I'm stuck
EDIT: Is there a way to finish it with induction?