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.
 A: 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 $\det A = 1/2$.
A: 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.
A: There is a formula for the characteristic polynomial for $3 \times 3$ matrices which says that 
$$
c_A(\lambda) = \det(A - \lambda I) = -\lambda^3 + c_2 \lambda^2 - c_1 \lambda + \det(A).
$$
(There is an analog for $2 \times 2$ matrices but I don't think there is a general formula for every $n$.) Since 
$$
2A^2 - A + I = 0,
$$
we know that the minimal polynomial $m_A(\lambda)$ satisfies
$$
m_A(\lambda) \, | \, 2\lambda^2 - \lambda + 1 = 2 \left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right) \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right).
$$
Therefore, the characteristic polynomial is among the following : 
$$
-\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right)^3 \\
-\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right)^2 \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right) \\
-\left( \lambda - \left( \frac{1 + \sqrt{-7}}4 \right) \right) \left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right)^2 \\
-\left( \lambda - \left( \frac{1 - \sqrt{-7}}4 \right) \right)^3.
$$
This gives you only four options for the determinant (I didn't compute them, it probably gives ugly complex numbers). I don't know how to distinguish the four possibilities though. Maybe $\det(A) \neq 0$ can help. Maybe.
Hope that helps,
A: As pointed out in other answers, no such real matrix exist. For complex matrix $A$, note  that $$I=2AA^{-1}=A(I-2A)=A-2A^2$$ Now let us assume $A$ is diagonal with all the diagonal entries equal to $a$, then each diagonal entry of $A-2A^2$ equals $a-2a^2$. Now from the equation $a-2a^2=1$, we have $a=(1\pm\sqrt{7}i)/4$. So for complex $A$ possible values of $\det (A)$ are $a^3.$
