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?