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.

Diagonalize the matrix first.
Let $A$ be your matrix, and you get $A = S\cdot J \cdot S^{-1}$, where 
$S = \begin{bmatrix}
1 & -a/b \\
1 & 1 \\
\end{bmatrix}$
 and 
$J = \begin{bmatrix}
1 & 0 \\
0 & 1-a-b \\
\end{bmatrix}$.
Then calculate $A^n = (S\cdot J \cdot S^{-1})^n = S \cdot J^n \cdot S^{-1}$ (just write things down and you will see this expression holds).
So the problem reduces to computing $J^n$, which is automatic since it's a diagonal matrix.
