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.

No such $A$ exists. Hence we cannot speak of its determinant.
Suppose $A$ is real and $A^{-1}=I-2A$. Then $A^2-\frac12A+\frac12I=0$. Hence the minimal polynomial $m_A$ of $A$ must divide $x^2-\frac12x+\frac12$, which has no real root. Therefore $m_A(x)=x^2-\frac12x+\frac12$. But the minimal polynomial and characteristic polynomial $p_A$ of $A$ must have identical irreducible factors, and this cannot happen because $p_A$ has degree 3 and $m_A$ is an irreducible polynomial of degree 2.
Edit: The OP says that the question appears on an extrance exam paper, and four answers are given: (a) $1/2$, (b) $−1/2$, (c) $1$, (d) $2$. It seems that there's a typo in the exam question and $A$ is probably 2x2. If this is really the case, then the above argument shows that the characteristic polynomial of $A$ is $x^2-\frac12x+\frac12$. Hence $\det A = 1/2$.
