We can invert a 2x2 matrix with 6 multiplies and one inversion. By Strassen's algorithm, we can multiply 2x2 matrices with 7 multiplies.
Partition your 4x4 matrix as
$$\left( \begin{matrix} A & B \\ C & D \end{matrix} \right) $$
We will reduce this to the identity.
Compute the inverse of $A$.
Perform Gaussian elimination
$$\left( \begin{matrix} I & A^{-1} B \\ 0 & D - CA^{-1}B \end{matrix} \right) $$
Let $E = D - C A^{-1} B$. Now compute $E^{-1}$. We can now finish Gaussian elimination.
Repeating these operations on an identity matrix tells us that the inverse is
$$ \left( \begin{matrix} A^{-1} + A^{-1} B E^{-1} C A^{-1} & -A^{-1} B E^{-1}
\\ -E^{-1} C A^{-1} & E^{-1} \end{matrix} \right) $$
The calculation of this inverse requires two matrix inversions (12 multiplies and 2 real inversions), and six 2x2 multiplies:
- $C A^{-1}$
- $(C A^{-1}) B$
- $E^{-1} (C A^{-1})$
- $A^{-1} B$
- $(A^{-1} B) E^{-1}$
- $(A^{-1} B)(E^{-1} C A^{-1})$
for 54 multiplies and 2 real inversions in all.
Of course, using Strassen's algorithm for 2x2 matrices is a terrible idea. I don't know if the above organization of the calculation is reasonable or not even if you don't use Strasses algorithm; I suspect it is unlikely.
If $A$ turns out to be singular, you have to partition the matrix differently to do the calculation.
Best Answer
Hint: Scaling a single row (or column) by a constant $k$ results in the determinant being scaled by $k$. So what happens if the entire matrix is scaled by $k$?