Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7 = I$, find an expression for $B^{-1}$ in terms of only $B$. Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7=I$, find an expression for $B^{-1}$ in terms of only $B$.  
I don't know where to start. Thanks in advance.

Since $B^2 + B^4 + B^7 = I$, we have $B(B + B^3 + B^6) = I$; thus $B^{-1} = B + B^3 + B^6$.  Note also that since $B^7 + B^4 + B^2 - I = 0$, any eigenvalue $\lambda$ of $B$ must satisfy $\lambda^7 + \lambda^4 + \lambda^2 - 1 =0$, hence cannot be $0$; thus the hypothesis that $B$ is invertable is extraneous, given that $B$ satisfies $B^2 + B^4 + B^7 = I$.  Hope this helps.
