Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:

Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
Solution: Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, by power series expansion, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A + 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, compute the product $BB^{-1}$, and you'll find that, miraculously, this answer works!

I discovered this solution technique some time ago while exploring a similar problem in Wolfram Mathematica. However, I have no idea why any of these manipulations should produce a meaningful answer when scalar and matrix inversion are such different operations. Why does this method work? Is there something deeper going on here than a serendipitous coincidence in series expansion coefficients?

The basic rule of thumb for such problems is that as long as all of the components of your equations are invertible and any multiplications used will commute, things will work just fine (this is the naive way of making the Banach Algebra comment).  Since all along we knew we were going to find out that $B$ was invertible (and since powers of $A$ are clearly invertible), you're safe to go ahead.
However, these sorts of problems often have easier solutions.  Hint: $B = A^2 - 2A + 2I = A^3 + A^2 - 2A = A(A^2 + A - 2I) = A(A+2I)(A-I)$.  Now finish with applications of the Neumann Lemma.
When these types of problems also give you the information that $A$ is, say, symmetric or normal, or has such-and-such eigenvalues, you should look to abuse the spectral theorem to get very quick proofs by proving for diagonal, then generally.
Exercise: If $A$ is real normal s.t. $A^3 = A^2 +A - I$, prove that $A$ is invertible - in fact, $A^2 = I$.
