Find the determinant of $A$ satisfying $A^{-1}=I-2A.$ I am stuck with the following problem:
Let $A$ be a $3\times 3$ matrix over real numbers satisfying $A^{-1}=I-2A.$
Then find the value of det$(A).$
I do not know how to proceed. Can someone point me in the right direction? Thanks in advance for your time.

Let $\lambda$ be a real eigenvalue with eigenvector $x$ (there is a real root to the characteristic equation). Since $A$ is invertible, $\lambda\neq 0$, so $Ax = \lambda x$ and $A^{-1}x=\lambda^{-1}x.$ Putting these into $A^{-1}x=x-2Ax$ gives $2\lambda^2-\lambda+1=0,$ contradicting that $\lambda$ is real. Hence no such $A$ exists.

It is now abundantly clear there was a typo in the question. I showed above no such real matrix exists. Even if we allow complex entries, the characteristic polynomial has the form $$(x^2-x/2+1/2)(x-z)$$ for some $z\in \mathbb{C}\setminus{\mathbb{R}}$ and the determinant is not real, so not one of the options.
