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.
 A: 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 $A^n$.
A: 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.
A: 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 $
