Calculating $B^{10}$ 
Calculate $B^{10}$ when $$B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$$


The way I did it was
$$ B = I + A $$
where
$$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
and $A^2=-I$. Since $A$ and $I$ are commutative,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= (I+A)(2A-2I) = -4I\\ \vdots \\ B^{10} &= 32A \end{aligned}$$
Is there a simpler method or a smarter approach if you want to do this for, e.g., $B^{100}$?

$$
B^2 = 2A, A^2=-I \implies
B^{100}= (B^2)^{50} = (2A)^{50} = 2^{50} A^{50} = 2^{50} (A^{2})^{25} = 2^{50} (-I)^{25} = -2^{50} I
$$
