For a given matrix B el. 12x12 the inverse is defined that $I=BB^{-1}$
In my example it is also given that $I + B^3 = 2B^2 - 4B$.
Does the inverse of $B$ exist and is there a definit answer or just a general?
And can you do the following steps?
$$ \begin{align}I &= -B^3 + 2B^2 - 4B \\I &= B\left(-B^2 + 2B - 4I\right) \end{align}$$ with $$ \begin{align} B^{-1} &=-B^2 + 2B - 4I \\ I &= -B\left(B^2 - 2B + I + 3I\right) \\ I &= -B \left( (B - I)^2 + 3I \right) \end{align} $$
and, $$ \begin{align}B^{-1} &=-(B-I)^2 - 3I \end{align}$$
Here is a partial answer not addressed in the comments.
If $B$ has entries over say $\mathbb{R}$, then you'll notice that the minimal polynomial of $B$ divides $x^3 -2x^2 + 4x + 1$, which does not have $0$ as a root. Since the characteristic polynomial has the same roots as the minimal polynomial, $B$ admits nonzero eigenvalues. Thus $B$ is invertible.
Edit: A WolframAlpha computation shows that $x^3 - 2x^2 + 4x + 1$ should have one real root and two complex roots. If $B$ is a $12\times 12$ real matrix, then the matrix whose diagonal is the root of $x^3 - 2x^2 + 4x + 1$ is a solution to the problem.