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.

Note that the characteristic polynomial of $A$ is given by
$$\begin{align}
\chi_A(\lambda) &= \lambda^2 - (2-a-b)\lambda + 1-a-b \\
&= \left( \lambda - \left( 1 - \frac{a+b}{2} \right) \right)^2 - \left( \frac{a+b}{2} \right)^2
\end{align}$$
so the eigenvalues of $A$ are $1-\dfrac{a+b}{2} \pm \dfrac{a+b}{2}$, i.e. $1$ and $1-a-b$.
Use these to find an invertible matrix $P$ such that $A = P \begin{pmatrix} 1 & 0 \\ 0 & 1-a-b \end{pmatrix} P^{-1}$; the columns of $P$ are eigenvectors of $A$.
When you've done this, you'll have
$$A^n = P \begin{pmatrix} 1 & 0 \\ 0 & 1-a-b \end{pmatrix}^n P^{-1} = P \begin{pmatrix} 1 & 0 \\ 0 & (1-a-b)^n \end{pmatrix} P^{-1}$$
and then expanding the matrix product gives a closed-form expression for $A^n$.
