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., $B^{100}$?
On
I would use the fact that$$B=\sqrt2\begin{bmatrix}\cos\left(\frac\pi4\right)&-\sin\left(\frac\pi4\right)\\\sin\left(\frac\pi4\right)&\cos\left(\frac\pi4\right)\end{bmatrix},$$and that therefore$$(\forall n\in\Bbb N):B^n=2^{n/2}\begin{bmatrix}\cos\left(\frac{n\pi}4\right)&-\sin\left(\frac{n\pi}4\right)\\\sin\left(\frac{n\pi}4\right)&\cos\left(\frac{n\pi}4\right)\end{bmatrix}.$$
On
If you use matrix representation of complex numbers: $$ a+bi \leftrightarrow\begin{pmatrix} a&-b\\b&a \end{pmatrix}\tag{1} $$ then using also the polar form of complex numbers gives: $$ w=(1+i)^n=(\sqrt{2}e^{i\pi/4})^n=(\sqrt{2})^n\exp(in\pi/4)=(\sqrt{2})^n(\cos(n\pi/4)+i\sin(n\pi/4)) $$
which by (1) can be translated into a matrix (see also José's answer).
Let $z=\sqrt{2}\omega$ with $\omega=e^{i\pi/4}$. Notice that $\omega^4=-1$. Then for any integer $k$ $$ z^{4k}=2^{2k}(-1)^k $$ (see also J.W. Tanner's answer).
$$ B^2 = 2A, A^2=-I \implies B^{100}= (B^2)^{50} = (2A)^{50} = 2^{50} A^{50} = 2^{50} (A^{2})^{25} = 2^{50} (-I)^{25} = -2^{50} I $$