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.

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,
