<s>[INST] Compute $\left[\begin{smallmatrix}1-a & a \\ b & 1-b\end{smallmatrix}\right]^n$ Compute $\begin{bmatrix}1-a & a \\ b & 1-b\end{bmatrix}^n$, where the power of $n\in\mathbb N$ denotes multiplying the matrix by itself $n$ times; $a,b\in[0,1]$.
Edit: 
I considered using induction, computed the desired matrix: 
$$\begin{bmatrix}(1-a)^2+ab & a(2-a-b) \\ b(2-a-b) & (1-b)^2+ab\end{bmatrix}$$ and $$\begin{bmatrix}(1-a)^3+ab(1-a)+ba(2-a-b) & a(1-a)^2+a^2b+a^2(2-a-b) \\ (1-a)b(2-a-b)+b(1-b)^2+ab^2 & ab(2-a-b)+(1-b)^3+ab(1-b)\end{bmatrix}$$ for $n=2$ and $3$, respectively. However, I fail to see a pattern here that can be used as an induction hypothesis.
 [/INST] Let $ n $ be a positive integer, $ a,b\in\left[0,1\right] $ such that $ a+b\neq 0 \cdot $
Denoting $ J=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right) $, observe that : $ \left(\begin{matrix}1-a&a\\b&1-b\end{matrix}\right)=I_{2}+J \cdot $
We have that: \begin{aligned} J^{2}=\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right)\left(\begin{matrix}-a&a\\b&-b\end{matrix}\right)=\left(\begin{matrix}a^{2}+ab&-a^{2}-ab\\-b^{2}-ab&b^{2}+ab\end{matrix}\right)&=-\left(a+b\right)\left(\begin{matrix}-a&a\\ b& -b\end{matrix}\right)\\&=-\left(a+b\right)J \end{aligned}
Meaning, $ \left(\forall k\in\mathbb{N}\right),\ J^{k+1}=\left(-1\right)^{k}\left(a+b\right)^{k}J \cdot $
Thus, $ \left(\begin{matrix}1-a&a\\b&1-b\end{matrix}\right)^{n}=\sum\limits_{k=0}^{n}{\binom{n}{k}J^{k}}=I_{2}+nJ+\sum\limits_{k=2}^{n}{\binom{n}{k}J^{k}} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I_{2}+nJ+\left(\sum\limits_{k=2}^{n}{\left(-1\right)^{k-1}\binom{n}{k}\left(a+b\right)^{k-1}}\right)J $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =I_{2}+\frac{1-\left(1-a-b\right)^{n}}{a+b}J $
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