Is there anything special about this matrix? I've just encountered a matrix which seems to display nothing special to me:
$$B=\begin{pmatrix}1&4&2\\0 &-3 &-2\\ 0 &4 &3 \end{pmatrix}$$
But further observation reveals something stunning:
$$B^n=\cases{{I}&n is even\\{B}&n is odd}$$
So it leads me to wonder if there is indeed some special properties of this matrix $B$, or more probably, $B$ belongs to a whole special class of matrices whose name I don't know? Could you drop me a hint? Thanks in advance.

EDIT I think I was being a bit stupid... It suddenly hit me that any matrix $B$ such that $B^2=I$ will have this property.

the eigenvalues of $B$ are given by $$0=\det\pmatrix{1-\lambda&4&2\\0&-3-\lambda&-2\\0&4&3-\lambda} =(1-\lambda)(\lambda^2 -1)=-(\lambda -1)^2(\lambda+1)$$  eigenvectors corresponding to the eigenvalue $1$ are $u = (1,0,0)^T, v = (0,1, -2)^T$ and an eigenvector corresponding to the eigenvalue $-1$ is $w=(1,-1,1).$ therefore the spectral decomposition of $B$ is 
$$B = \frac 1{u^Tu}uu^T+  \frac 1{v^Tv}vv^T- \frac 1{w^Tw}ww^T.$$ this implies that $$B^n = \begin{cases} I & \text{ if } n \text{ is even }\\
  B & \text{ if } n \text{ is odd }\end{cases} $$
